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受欢迎的三角函数 >

sin^4(x)=-1/8

解答

sin^{4} (x) = - \frac{1}{8}

解答

x \in R 无解

求解步骤

sin^{4} (x) = - \frac{1}{8}

用替代法求解

sin^{4} (x) = - \frac{1}{8}

令

: sin (x) = w

w^{4} = - \frac{1}{8}

w^{4} = - \frac{1}{8} : w = \frac{1}{2 4 2} + \frac{1}{2 4 2} i, w = - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}} - \frac{1}{2 4 2} i, w = - \frac{1}{2 ^{\frac{5}{4}}} + \frac{1}{2 4 2} i, w = \frac{1}{2 ^{\frac{5}{4}}} - \frac{1}{2 4 2} i

w^{4} = - \frac{1}{8}

用

u = w^{2}

和

u^{2} = w^{4}

改写方程式

u^{2} = - \frac{1}{8}

解

u^{2} = - \frac{1}{8} : u = i \frac{2}{4}, u = - i \frac{2}{4}

u^{2} = - \frac{1}{8}

对于

(g (x))^{2} = f (a)

解为

g (x) = f (a), - f (a)

u = - \frac{1}{8}, u = - - \frac{1}{8}

化简

- \frac{1}{8} : i \frac{2}{4}

- \frac{1}{8}

使用根式运算法则:

- a = - 1 a

- \frac{1}{8} = - 1 \frac{1}{8}

= - 1 \frac{1}{8}

使用虚数运算法则:

- 1 = i

= i \frac{1}{8}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

\frac{1}{8} = \frac{1}{8}

= i \frac{1}{8}

8 = 22

8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

= i \frac{1}{2 2}

使用法则

1 = 1

= i \frac{1}{2 2}

\frac{1}{2 2} = \frac{2}{4}

\frac{1}{2 2}

乘以共轭根式

\frac{2}{2}

= \frac{1 \cdot 2}{2 2 2}

1 \cdot 2 = 2

222 = 4

222

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

= 2^{1 + \frac{1}{2} + \frac{1}{2}}

同类项相加:

\frac{1}{2} + \frac{1}{2} = 2 \cdot \frac{1}{2}

= 2^{1 + 2 \cdot \frac{1}{2}}

2 \cdot \frac{1}{2} = 1

2 \cdot \frac{1}{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2^{1 + 1}

数字相加:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2}{4}

= - \frac{2}{4} i

u = i \frac{2}{4}, u = - i \frac{2}{4}

u = i \frac{2}{4}, u = - i \frac{2}{4}

代回

u = w^{2}

，求解

w

解

w^{2} = i \frac{2}{4} : w = \frac{1}{2 4 2} + \frac{1}{2 4 2} i, w = - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}} - \frac{1}{2 4 2} i

w^{2} = i \frac{2}{4}

替代

w = u + v i

(u + v i)^{2} = i \frac{2}{4}

展开

(u + v i)^{2} : (u^{2} - v^{2}) + 2 i uv

(u + v i)^{2}

= (u + i v)^{2}

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = u, b = v i

= u^{2} + 2 uv i + (v i)^{2}

(v i)^{2} = - v^{2}

(v i)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= i^{2} v^{2}

i^{2} = - 1

i^{2}

使用虚数运算法则:

i^{2} = - 1

= - 1

= (- 1) v^{2}

整理后得

= - v^{2}

= u^{2} + 2 i uv - v^{2}

将

u^{2} + 2 i uv - v^{2}

改写成标准复数形式:

(u^{2} - v^{2}) + 2 uv i

u^{2} + 2 i uv - v^{2}

将复数的实部和虚部分组

= (u^{2} - v^{2}) + 2 uv i

= (u^{2} - v^{2}) + 2 uv i

(u^{2} - v^{2}) + 2 i uv = i \frac{2}{4}

将

i \frac{2}{4}

改写成标准复数形式:

0 + \frac{2}{4} i

(u^{2} - v^{2}) + 2 i uv = 0 + \frac{2}{4} i

复数仅在实部和虚部均相等时才相等改写为方程组:

[u^{2} - v^{2} = 0 2 uv = \frac{2}{4}]

[u^{2} - v^{2} = 0 2 uv = \frac{2}{4}] : u = \frac{1}{2 4 2}, u = - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}}, v = \frac{1}{2 4 2} v = - \frac{1}{2 4 2}

[u^{2} - v^{2} = 0 2 uv = \frac{2}{4}]

对于

2 uv = \frac{2}{4}

将

u

移到一边:

u = \frac{1}{2 ^{\frac{5}{2}} v}

2 uv = \frac{2}{4}

因式分解数字:

4 = 2 \cdot 2

2 uv = \frac{2}{2 \cdot 2}

使用根式运算法则:

a = a a

2 = 22

2 uv = \frac{2}{2 2 \cdot 2}

约分:

2

2 uv = \frac{1}{2 \cdot 2}

2 uv = \frac{1}{2 2}

两边除以

2 v

2 uv = \frac{1}{2 2}

两边除以

2 v

\frac{2 uv}{2 v} = \frac{\frac{1}{2 2}}{2 v}

化简

\frac{2 uv}{2 v} = \frac{\frac{1}{2 2}}{2 v}

化简

\frac{2 uv}{2 v} : u

\frac{2 uv}{2 v}

约分:

2

= \frac{uv}{v}

约分:

v

= u

化简

\frac{\frac{1}{2 2}}{2 v} : \frac{1}{2 ^{\frac{5}{2}} v}

\frac{\frac{1}{2 2}}{2 v}

使用分式法则:

\frac{\frac{a}{b}}{c} = \frac{a}{b \cdot c}

= \frac{1}{2 2 \cdot 2 v}

化简

22 \cdot 2 v : 2^{\frac{5}{2}} v

22 \cdot 2 v

2 \cdot 2 = 2^{2}

2 \cdot 2

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2 = 2^{1 + 1}

= 2^{1 + 1}

数字相加:

1 + 1 = 2

= 2^{2}

= 2^{2} 2 v

使用根式运算法则:

a = a^{\frac{1}{2}}

2 = 2^{\frac{1}{2}}

= 2^{2} \cdot 2^{\frac{1}{2}} v

2^{2} \cdot 2^{\frac{1}{2}} = 2^{\frac{5}{2}}

2^{2} \cdot 2^{\frac{1}{2}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2^{2} \cdot 2^{\frac{1}{2}} = 2^{2 + \frac{1}{2}}

= 2^{2 + \frac{1}{2}}

2 + \frac{1}{2} = \frac{5}{2}

2 + \frac{1}{2}

将项转换为分式:

2 = \frac{2 \cdot 2}{2}

= \frac{2 \cdot 2}{2} + \frac{1}{2}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 \cdot 2 + 1}{2}

2 \cdot 2 + 1 = 5

2 \cdot 2 + 1

数字相乘:

2 \cdot 2 = 4

= 4 + 1

数字相加:

4 + 1 = 5

= 5

= \frac{5}{2}

= 2^{\frac{5}{2}}

= 2^{\frac{5}{2}} v

= \frac{1}{2 ^{\frac{5}{2}} v}

u = \frac{1}{2 ^{\frac{5}{2}} v}

u = \frac{1}{2 ^{\frac{5}{2}} v}

u = \frac{1}{2 ^{\frac{5}{2}} v}

将解

u = \frac{1}{2 ^{\frac{5}{2}} v}

代入

u^{2} - v^{2} = 0

对于

u^{2} - v^{2} = 0

，用

\frac{1}{2 ^{\frac{5}{2}} v}

替代

u : v = \frac{1}{2 4 2}, v = - \frac{1}{2 4 2}

对于

u^{2} - v^{2} = 0

，用

\frac{1}{2 ^{\frac{5}{2}} v}

替代

u

(\frac{1}{2 ^{\frac{5}{2}} v})^{2} - v^{2} = 0

解

(\frac{1}{2 ^{\frac{5}{2}} v})^{2} - v^{2} = 0 : v = \frac{1}{2 4 2}, v = - \frac{1}{2 4 2}

(\frac{1}{2 ^{\frac{5}{2}} v})^{2} - v^{2} = 0

化简

(\frac{1}{2 ^{\frac{5}{2}} v})^{2} : \frac{1}{32 v ^{2}}

(\frac{1}{2 ^{\frac{5}{2}} v})^{2}

\frac{1}{2 ^{\frac{5}{2}} v} = \frac{1}{2 ^{2} 2 v}

\frac{1}{2 ^{\frac{5}{2}} v}

2^{\frac{5}{2}} = 2^{2} 2

2^{\frac{5}{2}}

2^{\frac{5}{2}} = 2^{2 + \frac{1}{2}}

= 2^{2 + \frac{1}{2}}

使用指数法则:

x^{a + b} = x^{a} x^{b}

= 2^{2} \cdot 2^{\frac{1}{2}}

整理后得

= 2^{2} 2

= \frac{1}{2 ^{2} 2 v}

= (\frac{1}{2 ^{2} 2 v})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{( 2 ^{2} 2 v ) ^{2}}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

(2^{2} 2 v)^{2} = (2^{2})^{2} (2)^{2} v^{2}

= \frac{1 ^{2}}{( 2 ^{2} ) ^{2} ( 2 ) ^{2} v ^{2}}

(2^{2})^{2} : 2^{4}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 2^{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= 2^{4}

= \frac{1 ^{2}}{2 ^{4} ( 2 ) ^{2} v ^{2}}

(2)^{2} : 2

使用根式运算法则:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

= \frac{1 ^{2}}{2 ^{4} \cdot 2 v ^{2}}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{2 ^{4} \cdot 2 v ^{2}}

2^{4} \cdot 2 v^{2} = 2^{5} v^{2}

2^{4} \cdot 2 v^{2}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2^{4} \cdot 2 = 2^{4 + 1}

= 2^{4 + 1} v^{2}

数字相加:

4 + 1 = 5

= 2^{5} v^{2}

= \frac{1}{2 ^{5} v ^{2}}

2^{5} = 32

= \frac{1}{32 v ^{2}}

\frac{1}{32 v ^{2}} - v^{2} = 0

在两边乘以

32 v^{2}

\frac{1}{32 v ^{2}} - v^{2} = 0

在两边乘以

32 v^{2}

\frac{1}{32 v ^{2}} \cdot 32 v^{2} - v^{2} \cdot 32 v^{2} = 0 \cdot 32 v^{2}

化简

\frac{1}{32 v ^{2}} \cdot 32 v^{2} - v^{2} \cdot 32 v^{2} = 0 \cdot 32 v^{2}

化简

\frac{1}{32 v ^{2}} \cdot 32 v^{2} : 1

\frac{1}{32 v ^{2}} \cdot 32 v^{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 32 v ^{2}}{32 v ^{2}}

约分:

32

= \frac{1 \cdot v ^{2}}{v ^{2}}

约分:

v^{2}

= 1

化简

- v^{2} \cdot 32 v^{2} : - 32 v^{4}

- v^{2} \cdot 32 v^{2}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

v^{2} v^{2} = v^{2 + 2}

= - 32 v^{2 + 2}

数字相加:

2 + 2 = 4

= - 32 v^{4}

化简

0 \cdot 32 v^{2} : 0

0 \cdot 32 v^{2}

使用法则

0 \cdot a = 0

= 0

1 - 32 v^{4} = 0

1 - 32 v^{4} = 0

1 - 32 v^{4} = 0

解

1 - 32 v^{4} = 0 : v = \frac{1}{2 4 2}, v = - \frac{1}{2 4 2}

1 - 32 v^{4} = 0

将

1

到右边

1 - 32 v^{4} = 0

两边减去

1

1 - 32 v^{4} - 1 = 0 - 1

化简

- 32 v^{4} = - 1

- 32 v^{4} = - 1

两边除以

- 32

- 32 v^{4} = - 1

两边除以

- 32

\frac{- 32 v ^{4}}{- 32} = \frac{- 1}{- 32}

化简

v^{4} = \frac{1}{32}

v^{4} = \frac{1}{32}

对于

x^{n} = f (a)

，n 为偶数，解为

x = n f (a), - n f (a)

v = 4 \frac{1}{32}, v = - 4 \frac{1}{32}

4 \frac{1}{32} = \frac{1}{2 4 2}

4 \frac{1}{32}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b}, a \geq 0, b \geq 0

= \frac{4 1}{4 32}

使用根式运算法则:

n 1 = 1

41 = 1

= \frac{1}{4 32}

432 = 242

432

32

质因数分解:

2^{5}

32

32

除以

232 = 16 \cdot 2

= 2 \cdot 16

16

除以

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{5}

= 4 2^{5}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

2^{5} = 2^{4} \cdot 2

= 4 2^{4} \cdot 2

使用根式运算法则:

n ab = n a n b, a \geq 0, b \geq 0

4 2^{4} \cdot 2 = 4 2^{4} 42

= 4 2^{4} 42

使用根式运算法则:

n a^{n} = a, a \geq 0

4 2^{4} = 2

= 242

= \frac{1}{2 4 2}

- 4 \frac{1}{32} = - \frac{1}{2 4 2}

- 4 \frac{1}{32}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b}, a \geq 0, b \geq 0

= - \frac{4 1}{4 32}

使用根式运算法则:

n 1 = 1

41 = 1

= - \frac{1}{4 32}

432 = 242

432

32

质因数分解:

2^{5}

32

32

除以

232 = 16 \cdot 2

= 2 \cdot 16

16

除以

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{5}

= 4 2^{5}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

2^{5} = 2^{4} \cdot 2

= 4 2^{4} \cdot 2

使用根式运算法则:

n ab = n a n b, a \geq 0, b \geq 0

4 2^{4} \cdot 2 = 4 2^{4} 42

= 4 2^{4} 42

使用根式运算法则:

n a^{n} = a, a \geq 0

4 2^{4} = 2

= 242

= - \frac{1}{2 4 2}

v = \frac{1}{2 4 2}, v = - \frac{1}{2 4 2}

v = \frac{1}{2 4 2}, v = - \frac{1}{2 4 2}

验证解

找到无定义的点（奇点）:

v = 0

取

(\frac{1}{2 ^{\frac{5}{2}} v})^{2} - v^{2}

的分母，令其等于零

解

2^{\frac{5}{2}} v = 0 : v = 0

2^{\frac{5}{2}} v = 0

两边除以

2^{\frac{5}{2}}

2^{\frac{5}{2}} v = 0

两边除以

2^{\frac{5}{2}}

\frac{2 ^{\frac{5}{2}} v}{2 ^{\frac{5}{2}}} = \frac{0}{2 ^{\frac{5}{2}}}

化简

\frac{2 ^{\frac{5}{2}} v}{2 ^{\frac{5}{2}}} = \frac{0}{2 ^{\frac{5}{2}}}

化简

\frac{2 ^{\frac{5}{2}} v}{2 ^{\frac{5}{2}}} : v

\frac{2 ^{\frac{5}{2}} v}{2 ^{\frac{5}{2}}}

约分:

2^{\frac{5}{2}}

= v

化简

\frac{0}{2 ^{\frac{5}{2}}} : 0

\frac{0}{2 ^{\frac{5}{2}}}

使用法则

\frac{0}{a} = 0

a \neq = 0

= 0

v = 0

v = 0

v = 0

以下点无定义

v = 0

将不在定义域的点与解相综合:

v = \frac{1}{2 4 2}, v = - \frac{1}{2 4 2}

将解

v = \frac{1}{2 4 2}, v = - \frac{1}{2 4 2}

代入

2 uv = \frac{2}{4}

对于

2 uv = \frac{2}{4}

，用

\frac{1}{2 4 2}

替代

v : u = \frac{1}{2 4 2}

对于

2 uv = \frac{2}{4}

，用

\frac{1}{2 4 2}

替代

v

2 u \frac{1}{2 4 2} = \frac{2}{4}

解

2 u \frac{1}{2 4 2} = \frac{2}{4} : u = \frac{1}{2 4 2}

2 u \frac{1}{2 4 2} = \frac{2}{4}

因式分解数字:

4 = 2 \cdot 2

2 u \frac{1}{2 4 2} = \frac{2}{2 \cdot 2}

使用根式运算法则:

a = a a

2 = 22

2 u \frac{1}{2 4 2} = \frac{2}{2 2 \cdot 2}

约分:

2

2 u \frac{1}{2 4 2} = \frac{1}{2 \cdot 2}

2 u \frac{1}{2 4 2} = \frac{1}{2 2}

在两边乘以

242

2 u \frac{1}{2 4 2} = \frac{1}{2 2}

在两边乘以

242

2 u \frac{1}{2 4 2} \cdot 242 = \frac{1 \cdot 2 4 2}{2 2}

化简

2 u \frac{1}{2 4 2} \cdot 242 = \frac{1 \cdot 2 4 2}{2 2}

化简

2 u \frac{1}{2 4 2} \cdot 242 : 2 u

2 u \frac{1}{2 4 2} \cdot 242

2 \cdot 2 = 2^{2}

2 \cdot 2

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2 = 2^{1 + 1}

= 2^{1 + 1}

数字相加:

1 + 1 = 2

= 2^{2}

= 2^{2} u \frac{1}{2 4 2} 42

使用根式运算法则:

n a = a^{\frac{1}{n}}

42 = 2^{\frac{1}{4}}

= 2^{2} u \frac{1}{2 4 2} \cdot 2^{\frac{1}{4}}

2^{2} \cdot 2^{\frac{1}{4}} = 2^{\frac{9}{4}}

2^{2} \cdot 2^{\frac{1}{4}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2^{2} \cdot 2^{\frac{1}{4}} = 2^{2 + \frac{1}{4}}

= 2^{2 + \frac{1}{4}}

2 + \frac{1}{4} = \frac{9}{4}

2 + \frac{1}{4}

将项转换为分式:

2 = \frac{2 \cdot 4}{4}

= \frac{2 \cdot 4}{4} + \frac{1}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 \cdot 4 + 1}{4}

2 \cdot 4 + 1 = 9

2 \cdot 4 + 1

数字相乘:

2 \cdot 4 = 8

= 8 + 1

数字相加:

8 + 1 = 9

= 9

= \frac{9}{4}

= 2^{\frac{9}{4}}

= 2^{\frac{9}{4}} u \frac{1}{2 4 2}

使用分式法则:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 ^{\frac{9}{4}} u \cdot 1}{2 4 2}

使用法则:

a \cdot 1 = a

2^{\frac{9}{4}} u \cdot 1 = 2^{\frac{9}{4}} u

= \frac{2 ^{\frac{9}{4}} u}{2 4 2}

消掉

\frac{2 ^{\frac{9}{4}} u}{2 4 2} : 2 u

\frac{2 ^{\frac{9}{4}} u}{2 4 2}

\frac{2 ^{\frac{9}{4}}}{2 4 2} = 2

\frac{2 ^{\frac{9}{4}}}{2 4 2}

化简

\frac{2 ^{\frac{9}{4}}}{2} : 2^{\frac{5}{4}}

\frac{2 ^{\frac{9}{4}}}{2}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

= 2^{\frac{9}{4} - 1}

\frac{9}{4} - 1 = \frac{5}{4}

\frac{9}{4} - 1

将项转换为分式:

1 = \frac{1 \cdot 4}{4}

= - \frac{1 \cdot 4}{4} + \frac{9}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 1 \cdot 4 + 9}{4}

- 1 \cdot 4 + 9 = 5

- 1 \cdot 4 + 9

数字相乘:

1 \cdot 4 = 4

= - 4 + 9

数字相加/相减:

- 4 + 9 = 5

= 5

= \frac{5}{4}

= 2^{\frac{5}{4}}

= \frac{2 ^{\frac{5}{4}}}{4 2}

使用根式运算法则:

n a = a^{\frac{1}{n}}

42 = 2^{\frac{1}{4}}

= \frac{2 ^{\frac{5}{4}}}{2 ^{\frac{1}{4}}}

化简

\frac{2 ^{\frac{5}{4}}}{2 ^{\frac{1}{4}}} : 2

\frac{2 ^{\frac{5}{4}}}{2 ^{\frac{1}{4}}}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

= 2^{\frac{5}{4} - \frac{1}{4}}

\frac{5}{4} - \frac{1}{4} = 1

\frac{5}{4} - \frac{1}{4}

使用法则

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{5 - 1}{4}

数字相减:

5 - 1 = 4

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

= 2^{1}

使用指数法则:

a^{1} = a

= 2

= 2

= 2 u

= 2 u

化简

\frac{1 \cdot 2 4 2}{2 2} : \frac{1}{4 2}

\frac{1 \cdot 2 4 2}{2 2}

数字相乘:

1 \cdot 2 = 2

= \frac{2 4 2}{2 2}

约分:

2

= \frac{4 2}{2}

使用根式运算法则:

n a = a^{\frac{1}{n}}

42 = 2^{\frac{1}{4}}

= \frac{2 ^{\frac{1}{4}}}{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

2 = 2^{\frac{1}{2}}

= \frac{2 ^{\frac{1}{4}}}{2 ^{\frac{1}{2}}}

化简

\frac{2 ^{\frac{1}{4}}}{2 ^{\frac{1}{2}}} : \frac{1}{4 2}

\frac{2 ^{\frac{1}{4}}}{2 ^{\frac{1}{2}}}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

= \frac{1}{2 ^{\frac{1}{2} - \frac{1}{4}}}

\frac{1}{2} - \frac{1}{4} = \frac{1}{4}

\frac{1}{2} - \frac{1}{4}

2, 4

的最小公倍数:

4

2, 4

最小公倍数 (LCM)

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

将每个因子乘以它在

2

或

4

中出现的最多次数

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

4

对于

\frac{1}{2} :

将分母和分子乘以

2

\frac{1}{2} = \frac{1 \cdot 2}{2 \cdot 2} = \frac{2}{4}

= \frac{2}{4} - \frac{1}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 - 1}{4}

数字相减:

2 - 1 = 1

= \frac{1}{4}

= \frac{1}{2 ^{\frac{1}{4}}}

使用指数法则:

a^{\frac{m}{n}} = n a^{m}, a \geq 0

2^{\frac{1}{4}} = 42

= \frac{1}{4 2}

= \frac{1}{4 2}

2 u = \frac{1}{4 2}

2 u = \frac{1}{4 2}

2 u = \frac{1}{4 2}

两边除以

2

2 u = \frac{1}{4 2}

两边除以

2

\frac{2 u}{2} = \frac{\frac{1}{4 2}}{2}

化简

\frac{2 u}{2} = \frac{\frac{1}{4 2}}{2}

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

约分:

2

= u

化简

\frac{\frac{1}{4 2}}{2} : \frac{1}{2 4 2}

\frac{\frac{1}{4 2}}{2}

使用分式法则:

\frac{\frac{a}{b}}{c} = \frac{a}{b \cdot c}

= \frac{1}{4 2 \cdot 2}

= \frac{1}{2 4 2}

u = \frac{1}{2 4 2}

u = \frac{1}{2 4 2}

u = \frac{1}{2 4 2}

对于

2 uv = \frac{2}{4}

，用

- \frac{1}{2 4 2}

替代

v : u = - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}}

对于

2 uv = \frac{2}{4}

，用

- \frac{1}{2 4 2}

替代

v

2 u (- \frac{1}{2 4 2}) = \frac{2}{4}

解

2 u (- \frac{1}{2 4 2}) = \frac{2}{4} : u = - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}}

2 u (- \frac{1}{2 4 2}) = \frac{2}{4}

因式分解数字:

4 = 2 \cdot 2

2 u (- \frac{1}{2 4 2}) = \frac{2}{2 \cdot 2}

使用根式运算法则:

a = a a

2 = 22

2 u (- \frac{1}{2 4 2}) = \frac{2}{2 2 \cdot 2}

约分:

2

2 u (- \frac{1}{2 4 2}) = \frac{1}{2 \cdot 2}

2 u (- \frac{1}{2 4 2}) = \frac{1}{2 2}

两边除以

2 (- \frac{1}{2 4 2})

2 u (- \frac{1}{2 4 2}) = \frac{1}{2 2}

两边除以

2 (- \frac{1}{2 4 2})

\frac{2 u ( - \frac{1}{2 4 2} )}{2 ( - \frac{1}{2 4 2} )} = \frac{\frac{1}{2 2}}{2 ( - \frac{1}{2 4 2} )}

化简

\frac{2 u ( - \frac{1}{2 4 2} )}{2 ( - \frac{1}{2 4 2} )} = \frac{\frac{1}{2 2}}{2 ( - \frac{1}{2 4 2} )}

化简

\frac{2 u ( - \frac{1}{2 4 2} )}{2 ( - \frac{1}{2 4 2} )} : u

\frac{2 u ( - \frac{1}{2 4 2} )}{2 ( - \frac{1}{2 4 2} )}

化简

\frac{2 u ( - \frac{1}{2 4 2} )}{2 ( - \frac{1}{2 4 2} )} : \frac{- 2 u \frac{1}{2 4 2}}{- 2 \cdot \frac{1}{2 4 2}}

\frac{2 u ( - \frac{1}{2 4 2} )}{2 ( - \frac{1}{2 4 2} )}

使用法则:

a (- b) = - ab

2 u (- \frac{1}{2 4 2}) = - 2 u \frac{1}{2 4 2}

= \frac{- 2 u \frac{1}{2 4 2}}{2 ( - \frac{1}{2 4 2} )}

使用法则:

a (- b) = - ab

2 (- \frac{1}{2 4 2}) = - 2 \cdot \frac{1}{2 4 2}

= \frac{- 2 u \frac{1}{2 4 2}}{- 2 \cdot \frac{1}{2 4 2}}

= \frac{- 2 u \frac{1}{2 4 2}}{- 2 \cdot \frac{1}{2 4 2}}

约分:

- 2

= \frac{u \frac{1}{2 4 2}}{\frac{1}{2 4 2}}

约分:

\frac{1}{2 4 2}

= u

化简

\frac{\frac{1}{2 2}}{2 ( - \frac{1}{2 4 2} )} : - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}}

\frac{\frac{1}{2 2}}{2 ( - \frac{1}{2 4 2} )}

使用分式法则:

\frac{\frac{a}{b}}{c} = \frac{a}{b \cdot c}

= \frac{1}{2 2 \cdot 2 ( - \frac{1}{2 4 2} )}

使用法则:

a (- b) = - ab

22 \cdot 2 (- \frac{1}{2 4 2}) = - 22 \cdot 2 \cdot \frac{1}{2 4 2}

= \frac{1}{- 2 2 \cdot 2 \cdot \frac{1}{2 4 2}}

- 22 \cdot 2 \cdot \frac{1}{2 4 2} = - 2^{\frac{5}{2}} \cdot \frac{1}{2 4 2}

- 22 \cdot 2 \cdot \frac{1}{2 4 2}

2 \cdot 2 = 2^{2}

2 \cdot 2

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2 = 2^{1 + 1}

= 2^{1 + 1}

数字相加:

1 + 1 = 2

= 2^{2}

= - 2^{2} 2 \frac{1}{2 4 2}

使用根式运算法则:

a = a^{\frac{1}{2}}

2 = 2^{\frac{1}{2}}

= - 2^{2} \cdot 2^{\frac{1}{2}} \cdot \frac{1}{2 4 2}

2^{2} \cdot 2^{\frac{1}{2}} = 2^{\frac{5}{2}}

2^{2} \cdot 2^{\frac{1}{2}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2^{2} \cdot 2^{\frac{1}{2}} = 2^{2 + \frac{1}{2}}

= 2^{2 + \frac{1}{2}}

2 + \frac{1}{2} = \frac{5}{2}

2 + \frac{1}{2}

将项转换为分式:

2 = \frac{2 \cdot 2}{2}

= \frac{2 \cdot 2}{2} + \frac{1}{2}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 \cdot 2 + 1}{2}

2 \cdot 2 + 1 = 5

2 \cdot 2 + 1

数字相乘:

2 \cdot 2 = 4

= 4 + 1

数字相加:

4 + 1 = 5

= 5

= \frac{5}{2}

= 2^{\frac{5}{2}}

= - 2^{\frac{5}{2}} \cdot \frac{1}{2 4 2}

= \frac{1}{- 2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}}

u = - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}}

u = - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}}

u = - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}}

将解代入原方程进行验证

将它们代入

u^{2} - v^{2} = 0

检验解是否符合

去除与方程不符的解。

检验

u = - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}}, v = - \frac{1}{2 4 2}

的解:

真

u^{2} - v^{2} = 0

代入

u = - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}}, v = - \frac{1}{2 4 2}

(- \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}})^{2} - (- \frac{1}{2 4 2})^{2} = 0

整理后得

0 = 0

真

检验

u = \frac{1}{2 4 2}, v = \frac{1}{2 4 2}

的解:

真

u^{2} - v^{2} = 0

代入

u = \frac{1}{2 4 2}, v = \frac{1}{2 4 2}

(\frac{1}{2 4 2})^{2} - (\frac{1}{2 4 2})^{2} = 0

整理后得

0 = 0

真

将它们代入

2 uv = \frac{2}{4}

检验解是否符合

去除与方程不符的解。

检验

u = - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}}, v = - \frac{1}{2 4 2}

的解:

真

2 uv = \frac{2}{4}

代入

u = - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}}, v = - \frac{1}{2 4 2}

2 (- \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}}) (- \frac{1}{2 4 2}) = \frac{2}{4}

整理后得

\frac{2}{4} = \frac{2}{4}

真

检验

u = \frac{1}{2 4 2}, v = \frac{1}{2 4 2}

的解:

真

2 uv = \frac{2}{4}

代入

u = \frac{1}{2 4 2}, v = \frac{1}{2 4 2}

2 \cdot \frac{1}{2 4 2} \cdot \frac{1}{2 4 2} = \frac{2}{4}

整理后得

\frac{2}{4} = \frac{2}{4}

真

因而，

u^{2} - v^{2} = 0, 2 uv = \frac{2}{4}

最后的解是

u = \frac{1}{2 4 2}, u = - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}}, v = \frac{1}{2 4 2} v = - \frac{1}{2 4 2}

w = u + v i

代回

w = \frac{1}{2 4 2} + \frac{1}{2 4 2} i, w = - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}} - \frac{1}{2 4 2} i

解

w^{2} = - i \frac{2}{4} : w = - \frac{1}{2 ^{\frac{5}{4}}} + \frac{1}{2 4 2} i, w = \frac{1}{2 ^{\frac{5}{4}}} - \frac{1}{2 4 2} i

w^{2} = - i \frac{2}{4}

替代

w = u + v i

(u + v i)^{2} = - i \frac{2}{4}

展开

(u + v i)^{2} : (u^{2} - v^{2}) + 2 i uv

(u + v i)^{2}

= (u + i v)^{2}

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = u, b = v i

= u^{2} + 2 uv i + (v i)^{2}

(v i)^{2} = - v^{2}

(v i)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= i^{2} v^{2}

i^{2} = - 1

i^{2}

使用虚数运算法则:

i^{2} = - 1

= - 1

= (- 1) v^{2}

整理后得

= - v^{2}

= u^{2} + 2 i uv - v^{2}

将

u^{2} + 2 i uv - v^{2}

改写成标准复数形式:

(u^{2} - v^{2}) + 2 uv i

u^{2} + 2 i uv - v^{2}

将复数的实部和虚部分组

= (u^{2} - v^{2}) + 2 uv i

= (u^{2} - v^{2}) + 2 uv i

(u^{2} - v^{2}) + 2 i uv = - i \frac{2}{4}

将

- i \frac{2}{4}

改写成标准复数形式:

0 - \frac{2}{4} i

(u^{2} - v^{2}) + 2 i uv = 0 - \frac{2}{4} i

复数仅在实部和虚部均相等时才相等改写为方程组:

[u^{2} - v^{2} = 0 2 uv = - \frac{2}{4}]

[u^{2} - v^{2} = 0 2 uv = - \frac{2}{4}] : (u = - \frac{1}{2 ^{\frac{5}{4}}}, u = \frac{1}{2 ^{\frac{5}{4}}}, v = \frac{1}{2 4 2} v = - \frac{1}{2 4 2})

[u^{2} - v^{2} = 0 2 uv = - \frac{2}{4}]

对于

2 uv = - \frac{2}{4}

将

u

移到一边:

u = - \frac{1}{2 ^{\frac{5}{2}} v}

2 uv = - \frac{2}{4}

因式分解数字:

4 = 2 \cdot 2

2 uv = - \frac{2}{2 \cdot 2}

使用根式运算法则:

a = a a

2 = 22

2 uv = - \frac{2}{2 2 \cdot 2}

约分:

2

2 uv = - \frac{1}{2 \cdot 2}

2 uv = - \frac{1}{2 2}

两边除以

2 v

2 uv = - \frac{1}{2 2}

两边除以

2 v

\frac{2 uv}{2 v} = \frac{- \frac{1}{2 2}}{2 v}

化简

\frac{2 uv}{2 v} = \frac{- \frac{1}{2 2}}{2 v}

化简

\frac{2 uv}{2 v} : u

\frac{2 uv}{2 v}

约分:

2

= \frac{uv}{v}

约分:

v

= u

化简

\frac{- \frac{1}{2 2}}{2 v} : - \frac{1}{2 ^{\frac{5}{2}} v}

\frac{- \frac{1}{2 2}}{2 v}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{\frac{1}{2 2}}{2 v}

使用分式法则:

\frac{\frac{a}{b}}{c} = \frac{a}{b \cdot c}

\frac{\frac{1}{2 2}}{2 v} = \frac{1}{2 2 \cdot 2 v}

= - \frac{1}{2 2 \cdot 2 v}

化简

22 \cdot 2 v : 2^{\frac{5}{2}} v

22 \cdot 2 v

2 \cdot 2 = 2^{2}

2 \cdot 2

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2 = 2^{1 + 1}

= 2^{1 + 1}

数字相加:

1 + 1 = 2

= 2^{2}

= 2^{2} 2 v

使用根式运算法则:

a = a^{\frac{1}{2}}

2 = 2^{\frac{1}{2}}

= 2^{2} \cdot 2^{\frac{1}{2}} v

2^{2} \cdot 2^{\frac{1}{2}} = 2^{\frac{5}{2}}

2^{2} \cdot 2^{\frac{1}{2}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2^{2} \cdot 2^{\frac{1}{2}} = 2^{2 + \frac{1}{2}}

= 2^{2 + \frac{1}{2}}

2 + \frac{1}{2} = \frac{5}{2}

2 + \frac{1}{2}

将项转换为分式:

2 = \frac{2 \cdot 2}{2}

= \frac{2 \cdot 2}{2} + \frac{1}{2}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 \cdot 2 + 1}{2}

2 \cdot 2 + 1 = 5

2 \cdot 2 + 1

数字相乘:

2 \cdot 2 = 4

= 4 + 1

数字相加:

4 + 1 = 5

= 5

= \frac{5}{2}

= 2^{\frac{5}{2}}

= 2^{\frac{5}{2}} v

= - \frac{1}{2 ^{\frac{5}{2}} v}

u = - \frac{1}{2 ^{\frac{5}{2}} v}

u = - \frac{1}{2 ^{\frac{5}{2}} v}

u = - \frac{1}{2 ^{\frac{5}{2}} v}

将解

u = - \frac{1}{2 ^{\frac{5}{2}} v}

代入

u^{2} - v^{2} = 0

对于

u^{2} - v^{2} = 0

，用

- \frac{1}{2 ^{\frac{5}{2}} v}

替代

u : v = \frac{1}{2 4 2}, v = - \frac{1}{2 4 2}

对于

u^{2} - v^{2} = 0

，用

- \frac{1}{2 ^{\frac{5}{2}} v}

替代

u

(- \frac{1}{2 ^{\frac{5}{2}} v})^{2} - v^{2} = 0

解

(- \frac{1}{2 ^{\frac{5}{2}} v})^{2} - v^{2} = 0 : v = \frac{1}{2 4 2}, v = - \frac{1}{2 4 2}

(- \frac{1}{2 ^{\frac{5}{2}} v})^{2} - v^{2} = 0

化简

(- \frac{1}{2 ^{\frac{5}{2}} v})^{2} : \frac{1}{32 v ^{2}}

(- \frac{1}{2 ^{\frac{5}{2}} v})^{2}

\frac{1}{2 ^{\frac{5}{2}} v} = \frac{1}{2 ^{2} 2 v}

\frac{1}{2 ^{\frac{5}{2}} v}

2^{\frac{5}{2}} = 2^{2} 2

2^{\frac{5}{2}}

2^{\frac{5}{2}} = 2^{2 + \frac{1}{2}}

= 2^{2 + \frac{1}{2}}

使用指数法则:

x^{a + b} = x^{a} x^{b}

= 2^{2} \cdot 2^{\frac{1}{2}}

整理后得

= 2^{2} 2

= \frac{1}{2 ^{2} 2 v}

= (- \frac{1}{2 ^{2} 2 v})^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- \frac{1}{2 ^{2} 2 v})^{2} = (\frac{1}{2 ^{2} 2 v})^{2}

= (\frac{1}{2 ^{2} 2 v})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{( 2 ^{2} 2 v ) ^{2}}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

(2^{2} 2 v)^{2} = (2^{2})^{2} (2)^{2} v^{2}

= \frac{1 ^{2}}{( 2 ^{2} ) ^{2} ( 2 ) ^{2} v ^{2}}

(2^{2})^{2} : 2^{4}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 2^{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= 2^{4}

= \frac{1 ^{2}}{2 ^{4} ( 2 ) ^{2} v ^{2}}

(2)^{2} : 2

使用根式运算法则:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

= \frac{1 ^{2}}{2 ^{4} \cdot 2 v ^{2}}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{2 ^{4} \cdot 2 v ^{2}}

2^{4} \cdot 2 v^{2} = 2^{5} v^{2}

2^{4} \cdot 2 v^{2}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2^{4} \cdot 2 = 2^{4 + 1}

= 2^{4 + 1} v^{2}

数字相加:

4 + 1 = 5

= 2^{5} v^{2}

= \frac{1}{2 ^{5} v ^{2}}

2^{5} = 32

= \frac{1}{32 v ^{2}}

\frac{1}{32 v ^{2}} - v^{2} = 0

在两边乘以

32 v^{2}

\frac{1}{32 v ^{2}} - v^{2} = 0

在两边乘以

32 v^{2}

\frac{1}{32 v ^{2}} \cdot 32 v^{2} - v^{2} \cdot 32 v^{2} = 0 \cdot 32 v^{2}

化简

\frac{1}{32 v ^{2}} \cdot 32 v^{2} - v^{2} \cdot 32 v^{2} = 0 \cdot 32 v^{2}

化简

\frac{1}{32 v ^{2}} \cdot 32 v^{2} : 1

\frac{1}{32 v ^{2}} \cdot 32 v^{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 32 v ^{2}}{32 v ^{2}}

约分:

32

= \frac{1 \cdot v ^{2}}{v ^{2}}

约分:

v^{2}

= 1

化简

- v^{2} \cdot 32 v^{2} : - 32 v^{4}

- v^{2} \cdot 32 v^{2}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

v^{2} v^{2} = v^{2 + 2}

= - 32 v^{2 + 2}

数字相加:

2 + 2 = 4

= - 32 v^{4}

化简

0 \cdot 32 v^{2} : 0

0 \cdot 32 v^{2}

使用法则

0 \cdot a = 0

= 0

1 - 32 v^{4} = 0

1 - 32 v^{4} = 0

1 - 32 v^{4} = 0

解

1 - 32 v^{4} = 0 : v = \frac{1}{2 4 2}, v = - \frac{1}{2 4 2}

1 - 32 v^{4} = 0

将

1

到右边

1 - 32 v^{4} = 0

两边减去

1

1 - 32 v^{4} - 1 = 0 - 1

化简

- 32 v^{4} = - 1

- 32 v^{4} = - 1

两边除以

- 32

- 32 v^{4} = - 1

两边除以

- 32

\frac{- 32 v ^{4}}{- 32} = \frac{- 1}{- 32}

化简

v^{4} = \frac{1}{32}

v^{4} = \frac{1}{32}

对于

x^{n} = f (a)

，n 为偶数，解为

x = n f (a), - n f (a)

v = 4 \frac{1}{32}, v = - 4 \frac{1}{32}

4 \frac{1}{32} = \frac{1}{2 4 2}

4 \frac{1}{32}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b}, a \geq 0, b \geq 0

= \frac{4 1}{4 32}

使用根式运算法则:

n 1 = 1

41 = 1

= \frac{1}{4 32}

432 = 242

432

32

质因数分解:

2^{5}

32

32

除以

232 = 16 \cdot 2

= 2 \cdot 16

16

除以

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{5}

= 4 2^{5}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

2^{5} = 2^{4} \cdot 2

= 4 2^{4} \cdot 2

使用根式运算法则:

n ab = n a n b, a \geq 0, b \geq 0

4 2^{4} \cdot 2 = 4 2^{4} 42

= 4 2^{4} 42

使用根式运算法则:

n a^{n} = a, a \geq 0

4 2^{4} = 2

= 242

= \frac{1}{2 4 2}

- 4 \frac{1}{32} = - \frac{1}{2 4 2}

- 4 \frac{1}{32}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b}, a \geq 0, b \geq 0

= - \frac{4 1}{4 32}

使用根式运算法则:

n 1 = 1

41 = 1

= - \frac{1}{4 32}

432 = 242

432

32

质因数分解:

2^{5}

32

32

除以

232 = 16 \cdot 2

= 2 \cdot 16

16

除以

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{5}

= 4 2^{5}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

2^{5} = 2^{4} \cdot 2

= 4 2^{4} \cdot 2

使用根式运算法则:

n ab = n a n b, a \geq 0, b \geq 0

4 2^{4} \cdot 2 = 4 2^{4} 42

= 4 2^{4} 42

使用根式运算法则:

n a^{n} = a, a \geq 0

4 2^{4} = 2

= 242

= - \frac{1}{2 4 2}

v = \frac{1}{2 4 2}, v = - \frac{1}{2 4 2}

v = \frac{1}{2 4 2}, v = - \frac{1}{2 4 2}

验证解

找到无定义的点（奇点）:

v = 0

取

(- \frac{1}{2 ^{\frac{5}{2}} v})^{2} - v^{2}

的分母，令其等于零

解

2^{\frac{5}{2}} v = 0 : v = 0

2^{\frac{5}{2}} v = 0

两边除以

2^{\frac{5}{2}}

2^{\frac{5}{2}} v = 0

两边除以

2^{\frac{5}{2}}

\frac{2 ^{\frac{5}{2}} v}{2 ^{\frac{5}{2}}} = \frac{0}{2 ^{\frac{5}{2}}}

化简

\frac{2 ^{\frac{5}{2}} v}{2 ^{\frac{5}{2}}} = \frac{0}{2 ^{\frac{5}{2}}}

化简

\frac{2 ^{\frac{5}{2}} v}{2 ^{\frac{5}{2}}} : v

\frac{2 ^{\frac{5}{2}} v}{2 ^{\frac{5}{2}}}

约分:

2^{\frac{5}{2}}

= v

化简

\frac{0}{2 ^{\frac{5}{2}}} : 0

\frac{0}{2 ^{\frac{5}{2}}}

使用法则

\frac{0}{a} = 0

a \neq = 0

= 0

v = 0

v = 0

v = 0

以下点无定义

v = 0

将不在定义域的点与解相综合:

v = \frac{1}{2 4 2}, v = - \frac{1}{2 4 2}

将解

v = \frac{1}{2 4 2}, v = - \frac{1}{2 4 2}

代入

2 uv = - \frac{2}{4}

对于

2 uv = - \frac{2}{4}

，用

\frac{1}{2 4 2}

替代

v : u = - \frac{1}{2 ^{\frac{5}{4}}}

对于

2 uv = - \frac{2}{4}

，用

\frac{1}{2 4 2}

替代

v

2 u \frac{1}{2 4 2} = - \frac{2}{4}

解

2 u \frac{1}{2 4 2} = - \frac{2}{4} : u = - \frac{1}{2 ^{\frac{5}{4}}}

2 u \frac{1}{2 4 2} = - \frac{2}{4}

因式分解数字:

4 = 2 \cdot 2

2 u \frac{1}{2 4 2} = - \frac{2}{2 \cdot 2}

使用根式运算法则:

a = a a

2 = 22

2 u \frac{1}{2 4 2} = - \frac{2}{2 2 \cdot 2}

约分:

2

2 u \frac{1}{2 4 2} = - \frac{1}{2 \cdot 2}

2 u \frac{1}{2 4 2} = - \frac{1}{2 2}

在两边乘以

242

2 u \frac{1}{2 4 2} = - \frac{1}{2 2}

在两边乘以

242

2 u \frac{1}{2 4 2} \cdot 242 = (- \frac{1}{2 2}) \cdot 242

化简

2 u \frac{1}{2 4 2} \cdot 242 = (- \frac{1}{2 2}) \cdot 242

化简

2 u \frac{1}{2 4 2} \cdot 242 : 2 u

2 u \frac{1}{2 4 2} \cdot 242

2 \cdot 2 = 2^{2}

2 \cdot 2

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2 = 2^{1 + 1}

= 2^{1 + 1}

数字相加:

1 + 1 = 2

= 2^{2}

= 2^{2} u \frac{1}{2 4 2} 42

使用根式运算法则:

n a = a^{\frac{1}{n}}

42 = 2^{\frac{1}{4}}

= 2^{2} u \frac{1}{2 4 2} \cdot 2^{\frac{1}{4}}

2^{2} \cdot 2^{\frac{1}{4}} = 2^{\frac{9}{4}}

2^{2} \cdot 2^{\frac{1}{4}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2^{2} \cdot 2^{\frac{1}{4}} = 2^{2 + \frac{1}{4}}

= 2^{2 + \frac{1}{4}}

2 + \frac{1}{4} = \frac{9}{4}

2 + \frac{1}{4}

将项转换为分式:

2 = \frac{2 \cdot 4}{4}

= \frac{2 \cdot 4}{4} + \frac{1}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 \cdot 4 + 1}{4}

2 \cdot 4 + 1 = 9

2 \cdot 4 + 1

数字相乘:

2 \cdot 4 = 8

= 8 + 1

数字相加:

8 + 1 = 9

= 9

= \frac{9}{4}

= 2^{\frac{9}{4}}

= 2^{\frac{9}{4}} u \frac{1}{2 4 2}

使用分式法则:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 ^{\frac{9}{4}} u \cdot 1}{2 4 2}

使用法则:

a \cdot 1 = a

2^{\frac{9}{4}} u \cdot 1 = 2^{\frac{9}{4}} u

= \frac{2 ^{\frac{9}{4}} u}{2 4 2}

消掉

\frac{2 ^{\frac{9}{4}} u}{2 4 2} : 2 u

\frac{2 ^{\frac{9}{4}} u}{2 4 2}

\frac{2 ^{\frac{9}{4}}}{2 4 2} = 2

\frac{2 ^{\frac{9}{4}}}{2 4 2}

化简

\frac{2 ^{\frac{9}{4}}}{2} : 2^{\frac{5}{4}}

\frac{2 ^{\frac{9}{4}}}{2}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

= 2^{\frac{9}{4} - 1}

\frac{9}{4} - 1 = \frac{5}{4}

\frac{9}{4} - 1

将项转换为分式:

1 = \frac{1 \cdot 4}{4}

= - \frac{1 \cdot 4}{4} + \frac{9}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 1 \cdot 4 + 9}{4}

- 1 \cdot 4 + 9 = 5

- 1 \cdot 4 + 9

数字相乘:

1 \cdot 4 = 4

= - 4 + 9

数字相加/相减:

- 4 + 9 = 5

= 5

= \frac{5}{4}

= 2^{\frac{5}{4}}

= \frac{2 ^{\frac{5}{4}}}{4 2}

使用根式运算法则:

n a = a^{\frac{1}{n}}

42 = 2^{\frac{1}{4}}

= \frac{2 ^{\frac{5}{4}}}{2 ^{\frac{1}{4}}}

化简

\frac{2 ^{\frac{5}{4}}}{2 ^{\frac{1}{4}}} : 2

\frac{2 ^{\frac{5}{4}}}{2 ^{\frac{1}{4}}}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

= 2^{\frac{5}{4} - \frac{1}{4}}

\frac{5}{4} - \frac{1}{4} = 1

\frac{5}{4} - \frac{1}{4}

使用法则

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{5 - 1}{4}

数字相减:

5 - 1 = 4

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

= 2^{1}

使用指数法则:

a^{1} = a

= 2

= 2

= 2 u

= 2 u

化简

(- \frac{1}{2 2}) \cdot 242 : - \frac{1}{2} 42

(- \frac{1}{2 2}) \cdot 242

使用法则:

(- a) = - a

(- \frac{1}{2 2}) = - \frac{1}{2 2}

= - \frac{1}{2 2} \cdot 242

- \frac{1}{2 2} \cdot 242 = - \frac{1}{2} 42

- \frac{1}{2 2} \cdot 242

将

2

转换为分数

: \frac{2}{1}

2

将项转换为分式:

2 = \frac{2}{1}

= \frac{2}{1}

= - \frac{1}{2 2} \cdot \frac{2}{1} 42

使用分式法则:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

\frac{1}{2 2} \cdot \frac{2}{1} = \frac{1 \cdot 2}{2 2 \cdot 1}

= - \frac{1 \cdot 2}{2 2 \cdot 1} 42

\frac{1 \cdot 2}{2 2 \cdot 1} = \frac{1}{2}

\frac{1 \cdot 2}{2 2 \cdot 1}

\frac{1 \cdot 2}{2 2 \cdot 1} = \frac{2}{2 2}

\frac{1 \cdot 2}{2 2 \cdot 1}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{2 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{2}{2 2}

= \frac{2}{2 2}

约分:

2

= \frac{1}{2}

= - \frac{1}{2} 42

= - \frac{1}{2} 42

2 u = - \frac{1}{2} 42

2 u = - \frac{1}{2} 42

2 u = - \frac{1}{2} 42

两边除以

2

2 u = - \frac{1}{2} 42

两边除以

2

\frac{2 u}{2} = \frac{- \frac{1}{2} 4 2}{2}

化简

\frac{2 u}{2} = \frac{- \frac{1}{2} 4 2}{2}

化简

\frac{2 u}{2} : u

\frac{2 u}{2}

约分:

2

= u

化简

\frac{- \frac{1}{2} 4 2}{2} : - \frac{1}{2 ^{\frac{5}{4}}}

\frac{- \frac{1}{2} 4 2}{2}

使用根式运算法则:

a = n a n a \cdot \cdot \cdot n a (n t im es)

2 = 42424242

= \frac{- \frac{1}{2} 4 2}{4 2 4 2 4 2 4 2}

约分:

42

= \frac{- \frac{1}{2}}{4 2 4 2 4 2}

\frac{- \frac{1}{2}}{4 2 4 2 4 2} = - \frac{1}{2 ^{\frac{5}{4}}}

\frac{- \frac{1}{2}}{4 2 4 2 4 2}

使用指数法则:

aaa = a^{3}

424242 = (42)^{3}

= \frac{- \frac{1}{2}}{( 4 2 ) ^{3}}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{\frac{1}{2}}{( 4 2 ) ^{3}}

(42)^{3} = 2^{\frac{3}{4}}

(42)^{3}

使用根式运算法则:

n a = a^{\frac{1}{n}}

= (2^{\frac{1}{4}})^{3}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{4} \cdot 3}

\frac{1}{4} \cdot 3 = \frac{3}{4}

\frac{1}{4} \cdot 3

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 3}{4}

数字相乘:

1 \cdot 3 = 3

= \frac{3}{4}

= 2^{\frac{3}{4}}

= - \frac{\frac{1}{2}}{2 ^{\frac{3}{4}}}

- \frac{\frac{1}{2}}{2 ^{\frac{3}{4}}} = - \frac{1}{2 ^{\frac{5}{4}}}

- \frac{\frac{1}{2}}{2 ^{\frac{3}{4}}}

使用分式法则:

\frac{\frac{a}{b}}{c} = \frac{a}{b \cdot c}

\frac{\frac{1}{2}}{2 ^{\frac{3}{4}}} = \frac{1}{2 \cdot 2 ^{\frac{3}{4}}}

= - \frac{1}{2 \cdot 2 ^{\frac{3}{4}}}

2 \cdot 2^{\frac{3}{4}} = 2^{\frac{5}{4}}

2 \cdot 2^{\frac{3}{4}}

使用根式运算法则:

a = a^{\frac{1}{2}}

2 = 2^{\frac{1}{2}}

= 2^{\frac{1}{2}} \cdot 2^{\frac{3}{4}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2^{\frac{1}{2}} \cdot 2^{\frac{3}{4}} = 2^{\frac{1}{2} + \frac{3}{4}}

= 2^{\frac{1}{2} + \frac{3}{4}}

\frac{1}{2} + \frac{3}{4} = \frac{5}{4}

\frac{1}{2} + \frac{3}{4}

2, 4

的最小公倍数:

4

2, 4

最小公倍数 (LCM)

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

将每个因子乘以它在

2

或

4

中出现的最多次数

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

4

对于

\frac{1}{2} :

将分母和分子乘以

2

\frac{1}{2} = \frac{1 \cdot 2}{2 \cdot 2} = \frac{2}{4}

= \frac{2}{4} + \frac{3}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 + 3}{4}

数字相加:

2 + 3 = 5

= \frac{5}{4}

= 2^{\frac{5}{4}}

= - \frac{1}{2 ^{\frac{5}{4}}}

= - \frac{1}{2 ^{\frac{5}{4}}}

= - \frac{1}{2 ^{\frac{5}{4}}}

u = - \frac{1}{2 ^{\frac{5}{4}}}

u = - \frac{1}{2 ^{\frac{5}{4}}}

u = - \frac{1}{2 ^{\frac{5}{4}}}

对于

2 uv = - \frac{2}{4}

，用

- \frac{1}{2 4 2}

替代

v : u = \frac{1}{2 ^{\frac{5}{4}}}

对于

2 uv = - \frac{2}{4}

，用

- \frac{1}{2 4 2}

替代

v

2 u (- \frac{1}{2 4 2}) = - \frac{2}{4}

解

2 u (- \frac{1}{2 4 2}) = - \frac{2}{4} : u = \frac{1}{2 ^{\frac{5}{4}}}

2 u (- \frac{1}{2 4 2}) = - \frac{2}{4}

因式分解数字:

4 = 2 \cdot 2

2 u (- \frac{1}{2 4 2}) = - \frac{2}{2 \cdot 2}

使用根式运算法则:

a = a a

2 = 22

2 u (- \frac{1}{2 4 2}) = - \frac{2}{2 2 \cdot 2}

约分:

2

2 u (- \frac{1}{2 4 2}) = - \frac{1}{2 \cdot 2}

2 u (- \frac{1}{2 4 2}) = - \frac{1}{2 2}

两边除以

2 (- \frac{1}{2 4 2})

2 u (- \frac{1}{2 4 2}) = - \frac{1}{2 2}

两边除以

2 (- \frac{1}{2 4 2})

\frac{2 u ( - \frac{1}{2 4 2} )}{2 ( - \frac{1}{2 4 2} )} = \frac{- \frac{1}{2 2}}{2 ( - \frac{1}{2 4 2} )}

化简

\frac{2 u ( - \frac{1}{2 4 2} )}{2 ( - \frac{1}{2 4 2} )} = \frac{- \frac{1}{2 2}}{2 ( - \frac{1}{2 4 2} )}

化简

\frac{2 u ( - \frac{1}{2 4 2} )}{2 ( - \frac{1}{2 4 2} )} : u

\frac{2 u ( - \frac{1}{2 4 2} )}{2 ( - \frac{1}{2 4 2} )}

化简

\frac{2 u ( - \frac{1}{2 4 2} )}{2 ( - \frac{1}{2 4 2} )} : \frac{- 2 u \frac{1}{2 4 2}}{- 2 \cdot \frac{1}{2 4 2}}

\frac{2 u ( - \frac{1}{2 4 2} )}{2 ( - \frac{1}{2 4 2} )}

使用法则:

a (- b) = - ab

2 u (- \frac{1}{2 4 2}) = - 2 u \frac{1}{2 4 2}

= \frac{- 2 u \frac{1}{2 4 2}}{2 ( - \frac{1}{2 4 2} )}

使用法则:

a (- b) = - ab

2 (- \frac{1}{2 4 2}) = - 2 \cdot \frac{1}{2 4 2}

= \frac{- 2 u \frac{1}{2 4 2}}{- 2 \cdot \frac{1}{2 4 2}}

= \frac{- 2 u \frac{1}{2 4 2}}{- 2 \cdot \frac{1}{2 4 2}}

约分:

- 2

= \frac{u \frac{1}{2 4 2}}{\frac{1}{2 4 2}}

约分:

\frac{1}{2 4 2}

= u

化简

\frac{- \frac{1}{2 2}}{2 ( - \frac{1}{2 4 2} )} : \frac{1}{2 ^{\frac{5}{4}}}

\frac{- \frac{1}{2 2}}{2 ( - \frac{1}{2 4 2} )}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{\frac{1}{2 2}}{2 ( - \frac{1}{2 4 2} )}

使用法则:

a (- b) = - ab

2 (- \frac{1}{2 4 2}) = - 2 \cdot \frac{1}{2 4 2}

= - \frac{\frac{1}{2 2}}{- 2 \cdot \frac{1}{2 4 2}}

- 2 \cdot \frac{1}{2 4 2} = - \frac{1}{4 2}

- 2 \cdot \frac{1}{2 4 2}

将

2

转换为分数

: \frac{2}{1}

2

将项转换为分式:

2 = \frac{2}{1}

= \frac{2}{1}

= - \frac{2}{1} \cdot \frac{1}{2 4 2}

使用分式法则:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

\frac{2}{1} \cdot \frac{1}{2 4 2} = \frac{2 \cdot 1}{1 \cdot 2 4 2}

= - \frac{2 \cdot 1}{1 \cdot 2 4 2}

\frac{2 \cdot 1}{1 \cdot 2 4 2} = \frac{1}{4 2}

\frac{2 \cdot 1}{1 \cdot 2 4 2}

\frac{2 \cdot 1}{1 \cdot 2 4 2} = \frac{2}{2 4 2}

\frac{2 \cdot 1}{1 \cdot 2 4 2}

数字相乘:

2 \cdot 1 = 2

= \frac{2}{1 \cdot 2 4 2}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{2 4 2}

= \frac{2}{2 4 2}

约分:

2

= \frac{1}{4 2}

= - \frac{1}{4 2}

= - \frac{\frac{1}{2 2}}{- \frac{1}{4 2}}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

\frac{\frac{1}{2 2}}{- \frac{1}{4 2}} = - \frac{\frac{1}{2 2}}{\frac{1}{4 2}}

= - (- \frac{\frac{1}{2 2}}{\frac{1}{4 2}})

使用法则:

- (- a) = a

- (- \frac{\frac{1}{2 2}}{\frac{1}{4 2}}) = \frac{\frac{1}{2 2}}{\frac{1}{4 2}}

= \frac{\frac{1}{2 2}}{\frac{1}{4 2}}

使用分式法则:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{1 \cdot 4 2}{2 2 \cdot 1}

消掉

\frac{1 \cdot 4 2}{2 2 \cdot 1} : \frac{1}{2 ^{\frac{5}{4}}}

\frac{1 \cdot 4 2}{2 2 \cdot 1}

\frac{1 \cdot 4 2}{2 2 \cdot 1} = \frac{4 2}{2 2}

\frac{1 \cdot 4 2}{2 2 \cdot 1}

使用法则:

1 \cdot a = a

1 \cdot 42 = 42

= \frac{4 2}{2 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{4 2}{2 2}

= \frac{4 2}{2 2}

使用根式运算法则:

a = n a n a \cdot \cdot \cdot n a (n t im es)

2 = 42424242

= \frac{4 2}{4 2 4 2 4 2 4 2 2}

约分:

42

= \frac{1}{4 2 4 2 4 2 2}

\frac{1}{4 2 4 2 4 2 2} = \frac{1}{2 ^{\frac{5}{4}}}

\frac{1}{4 2 4 2 4 2 2}

使用指数法则:

aaa = a^{3}

424242 = (42)^{3}

= \frac{1}{( 4 2 ) ^{3} 2}

(42)^{3} = 2^{\frac{3}{4}}

(42)^{3}

使用根式运算法则:

n a = a^{\frac{1}{n}}

= (2^{\frac{1}{4}})^{3}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{4} \cdot 3}

\frac{1}{4} \cdot 3 = \frac{3}{4}

\frac{1}{4} \cdot 3

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 3}{4}

数字相乘:

1 \cdot 3 = 3

= \frac{3}{4}

= 2^{\frac{3}{4}}

= \frac{1}{2 ^{\frac{3}{4}} 2}

2^{\frac{3}{4}} 2 = 2^{\frac{5}{4}}

2^{\frac{3}{4}} 2

使用根式运算法则:

a = a^{\frac{1}{2}}

2 = 2^{\frac{1}{2}}

= 2^{\frac{3}{4}} \cdot 2^{\frac{1}{2}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2^{\frac{3}{4}} \cdot 2^{\frac{1}{2}} = 2^{\frac{3}{4} + \frac{1}{2}}

= 2^{\frac{3}{4} + \frac{1}{2}}

\frac{3}{4} + \frac{1}{2} = \frac{5}{4}

\frac{3}{4} + \frac{1}{2}

4, 2

的最小公倍数:

4

4, 2

最小公倍数 (LCM)

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

将每个因子乘以它在

4

或

2

中出现的最多次数

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

4

对于

\frac{1}{2} :

将分母和分子乘以

2

\frac{1}{2} = \frac{1 \cdot 2}{2 \cdot 2} = \frac{2}{4}

= \frac{3}{4} + \frac{2}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{3 + 2}{4}

数字相加:

3 + 2 = 5

= \frac{5}{4}

= 2^{\frac{5}{4}}

= \frac{1}{2 ^{\frac{5}{4}}}

= \frac{1}{2 ^{\frac{5}{4}}}

= \frac{1}{2 ^{\frac{5}{4}}}

u = \frac{1}{2 ^{\frac{5}{4}}}

u = \frac{1}{2 ^{\frac{5}{4}}}

u = \frac{1}{2 ^{\frac{5}{4}}}

将解代入原方程进行验证

将它们代入

u^{2} - v^{2} = 0

检验解是否符合

去除与方程不符的解。

检验

u = \frac{1}{2 ^{\frac{5}{4}}}, v = - \frac{1}{2 4 2}

的解:

真

u^{2} - v^{2} = 0

代入

u = \frac{1}{2 ^{\frac{5}{4}}}, v = - \frac{1}{2 4 2}

(\frac{1}{2 ^{\frac{5}{4}}})^{2} - (- \frac{1}{2 4 2})^{2} = 0

整理后得

0 = 0

真

检验

u = - \frac{1}{2 ^{\frac{5}{4}}}, v = \frac{1}{2 4 2}

的解:

真

u^{2} - v^{2} = 0

代入

u = - \frac{1}{2 ^{\frac{5}{4}}}, v = \frac{1}{2 4 2}

(- \frac{1}{2 ^{\frac{5}{4}}})^{2} - (\frac{1}{2 4 2})^{2} = 0

整理后得

0 = 0

真

将它们代入

2 uv = - \frac{2}{4}

检验解是否符合

去除与方程不符的解。

检验

u = \frac{1}{2 ^{\frac{5}{4}}}, v = - \frac{1}{2 4 2}

的解:

真

2 uv = - \frac{2}{4}

代入

u = \frac{1}{2 ^{\frac{5}{4}}}, v = - \frac{1}{2 4 2}

2 \cdot \frac{1}{2 ^{\frac{5}{4}}} (- \frac{1}{2 4 2}) = - \frac{2}{4}

整理后得

- \frac{2}{4} = - \frac{2}{4}

真

检验

u = - \frac{1}{2 ^{\frac{5}{4}}}, v = \frac{1}{2 4 2}

的解:

真

2 uv = - \frac{2}{4}

代入

u = - \frac{1}{2 ^{\frac{5}{4}}}, v = \frac{1}{2 4 2}

2 (- \frac{1}{2 ^{\frac{5}{4}}}) \frac{1}{2 4 2} = - \frac{2}{4}

整理后得

- \frac{2}{4} = - \frac{2}{4}

真

因而，

u^{2} - v^{2} = 0, 2 uv = - \frac{2}{4}

最后的解是

(u = - \frac{1}{2 ^{\frac{5}{4}}}, u = \frac{1}{2 ^{\frac{5}{4}}}, v = \frac{1}{2 4 2} v = - \frac{1}{2 4 2})

w = u + v i

代回

w = - \frac{1}{2 ^{\frac{5}{4}}} + \frac{1}{2 4 2} i, w = \frac{1}{2 ^{\frac{5}{4}}} - \frac{1}{2 4 2} i

解为

w = \frac{1}{2 4 2} + \frac{1}{2 4 2} i, w = - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}} - \frac{1}{2 4 2} i, w = - \frac{1}{2 ^{\frac{5}{4}}} + \frac{1}{2 4 2} i, w = \frac{1}{2 ^{\frac{5}{4}}} - \frac{1}{2 4 2} i

w = sin (x)

代回

sin (x) = \frac{1}{2 4 2} + \frac{1}{2 4 2} i, sin (x) = - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}} - \frac{1}{2 4 2} i, sin (x) = - \frac{1}{2 ^{\frac{5}{4}}} + \frac{1}{2 4 2} i, sin (x) = \frac{1}{2 ^{\frac{5}{4}}} - \frac{1}{2 4 2} i

sin (x) = \frac{1}{2 4 2} + \frac{1}{2 4 2} i, sin (x) = - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}} - \frac{1}{2 4 2} i, sin (x) = - \frac{1}{2 ^{\frac{5}{4}}} + \frac{1}{2 4 2} i, sin (x) = \frac{1}{2 ^{\frac{5}{4}}} - \frac{1}{2 4 2} i

sin (x) = \frac{1}{2 4 2} + \frac{1}{2 4 2} i :

无解

sin (x) = \frac{1}{2 4 2} + \frac{1}{2 4 2} i

化简

\frac{1}{2 4 2} + \frac{1}{2 4 2} i : \frac{2 ^{\frac{3}{4}}}{4} + i \frac{2 ^{\frac{3}{4}}}{4}

\frac{1}{2 4 2} + \frac{1}{2 4 2} i

乘

\frac{1}{2 4 2} i : \frac{i}{2 4 2}

\frac{1}{2 4 2} i

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 i}{2 4 2}

乘以:

1 i = i

= \frac{i}{2 4 2}

= \frac{1}{2 4 2} + \frac{i}{2 4 2}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 + i}{2 4 2}

\frac{1 + i}{2 4 2}

有理化:

\frac{2 ^{\frac{3}{4}} ( 1 + i )}{4}

\frac{1 + i}{2 4 2}

乘以共轭根式

\frac{2 ^{\frac{3}{4}}}{2 ^{\frac{3}{4}}}

= \frac{( 1 + i ) \cdot 2 ^{\frac{3}{4}}}{2 4 2 \cdot 2 ^{\frac{3}{4}}}

242 \cdot 2^{\frac{3}{4}} = 4

242 \cdot 2^{\frac{3}{4}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2^{\frac{3}{4}} 42 = 2 \cdot 2^{\frac{3}{4}} \cdot 2^{\frac{1}{4}} = 2^{1 + \frac{3}{4} + \frac{1}{4}}

= 2^{1 + \frac{3}{4} + \frac{1}{4}}

化简

1 + \frac{3}{4} + \frac{1}{4} : 2

1 + \frac{3}{4} + \frac{1}{4}

将项转换为分式:

1 = \frac{1}{1}

= \frac{1}{1} + \frac{3}{4} + \frac{1}{4}

1, 4, 4

的最小公倍数:

4

1, 4, 4

最小公倍数 (LCM)

1

质因数分解

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

计算出由至少在以下一个数字中出现的因数组成的数字:

1, 4, 4

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

4

对于

\frac{1}{1} :

将分母和分子乘以

4

\frac{1}{1} = \frac{1 \cdot 4}{1 \cdot 4} = \frac{4}{4}

= \frac{4}{4} + \frac{3}{4} + \frac{1}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{4 + 3 + 1}{4}

数字相加:

4 + 3 + 1 = 8

= \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2 ^{\frac{3}{4}} ( 1 + i )}{4}

= \frac{2 ^{\frac{3}{4}} ( 1 + i )}{4}

将

\frac{2 ^{\frac{3}{4}} ( 1 + i )}{4}

改写成标准复数形式:

\frac{2 ^{\frac{3}{4}}}{4} + \frac{2 ^{\frac{3}{4}}}{4} i

\frac{2 ^{\frac{3}{4}} ( 1 + i )}{4}

分解

4 : 2^{2}

因式分解

4 = 2^{2}

= \frac{2 ^{\frac{3}{4}} ( 1 + i )}{2 ^{2}}

消掉

\frac{2 ^{\frac{3}{4}} ( 1 + i )}{2 ^{2}} : \frac{1 + i}{2 ^{\frac{5}{4}}}

\frac{2 ^{\frac{3}{4}} ( 1 + i )}{2 ^{2}}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{\frac{3}{4}}}{2 ^{2}} = \frac{1}{2 ^{2 - \frac{3}{4}}}

= \frac{1 + i}{2 ^{2 - \frac{3}{4}}}

数字相减:

2 - \frac{3}{4} = \frac{5}{4}

= \frac{1 + i}{2 ^{\frac{5}{4}}}

= \frac{1 + i}{2 ^{\frac{5}{4}}}

2^{\frac{5}{4}} = 242

2^{\frac{5}{4}}

2^{\frac{5}{4}} = 2^{1 + \frac{1}{4}}

= 2^{1 + \frac{1}{4}}

使用指数法则:

x^{a + b} = x^{a} x^{b}

= 2^{1} \cdot 2^{\frac{1}{4}}

整理后得

= 242

= \frac{1 + i}{2 4 2}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{1 + i}{2 4 2} = \frac{1}{2 4 2} + \frac{i}{2 4 2}

= \frac{1}{2 4 2} + \frac{i}{2 4 2}

\frac{1}{2 4 2} = \frac{2 ^{\frac{3}{4}}}{4}

\frac{1}{2 4 2}

乘以共轭根式

\frac{2 ^{\frac{3}{4}}}{2 ^{\frac{3}{4}}}

= \frac{1 \cdot 2 ^{\frac{3}{4}}}{2 4 2 \cdot 2 ^{\frac{3}{4}}}

1 \cdot 2^{\frac{3}{4}} = 2^{\frac{3}{4}}

242 \cdot 2^{\frac{3}{4}} = 4

242 \cdot 2^{\frac{3}{4}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2^{\frac{3}{4}} 42 = 2 \cdot 2^{\frac{3}{4}} \cdot 2^{\frac{1}{4}} = 2^{1 + \frac{3}{4} + \frac{1}{4}}

= 2^{1 + \frac{3}{4} + \frac{1}{4}}

化简

1 + \frac{3}{4} + \frac{1}{4} : 2

1 + \frac{3}{4} + \frac{1}{4}

将项转换为分式:

1 = \frac{1}{1}

= \frac{1}{1} + \frac{3}{4} + \frac{1}{4}

1, 4, 4

的最小公倍数:

4

1, 4, 4

最小公倍数 (LCM)

1

质因数分解

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

计算出由至少在以下一个数字中出现的因数组成的数字:

1, 4, 4

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

4

对于

\frac{1}{1} :

将分母和分子乘以

4

\frac{1}{1} = \frac{1 \cdot 4}{1 \cdot 4} = \frac{4}{4}

= \frac{4}{4} + \frac{3}{4} + \frac{1}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{4 + 3 + 1}{4}

数字相加:

4 + 3 + 1 = 8

= \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2 ^{\frac{3}{4}}}{4}

= \frac{1}{2 4 2} + \frac{2 ^{\frac{3}{4}}}{4} i

\frac{1}{2 4 2} = \frac{2 ^{\frac{3}{4}}}{4}

\frac{1}{2 4 2}

乘以共轭根式

\frac{2 ^{\frac{3}{4}}}{2 ^{\frac{3}{4}}}

= \frac{1 \cdot 2 ^{\frac{3}{4}}}{2 4 2 \cdot 2 ^{\frac{3}{4}}}

1 \cdot 2^{\frac{3}{4}} = 2^{\frac{3}{4}}

242 \cdot 2^{\frac{3}{4}} = 4

242 \cdot 2^{\frac{3}{4}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2^{\frac{3}{4}} 42 = 2 \cdot 2^{\frac{3}{4}} \cdot 2^{\frac{1}{4}} = 2^{1 + \frac{3}{4} + \frac{1}{4}}

= 2^{1 + \frac{3}{4} + \frac{1}{4}}

化简

1 + \frac{3}{4} + \frac{1}{4} : 2

1 + \frac{3}{4} + \frac{1}{4}

将项转换为分式:

1 = \frac{1}{1}

= \frac{1}{1} + \frac{3}{4} + \frac{1}{4}

1, 4, 4

的最小公倍数:

4

1, 4, 4

最小公倍数 (LCM)

1

质因数分解

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

计算出由至少在以下一个数字中出现的因数组成的数字:

1, 4, 4

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

4

对于

\frac{1}{1} :

将分母和分子乘以

4

\frac{1}{1} = \frac{1 \cdot 4}{1 \cdot 4} = \frac{4}{4}

= \frac{4}{4} + \frac{3}{4} + \frac{1}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{4 + 3 + 1}{4}

数字相加:

4 + 3 + 1 = 8

= \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2 ^{\frac{3}{4}}}{4}

= \frac{2 ^{\frac{3}{4}}}{4} + \frac{2 ^{\frac{3}{4}}}{4} i

= \frac{2 ^{\frac{3}{4}}}{4} + \frac{2 ^{\frac{3}{4}}}{4} i

无解

sin (x) = - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}} - \frac{1}{2 4 2} i :

无解

sin (x) = - \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}} - \frac{1}{2 4 2} i

化简

- \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}} - \frac{1}{2 4 2} i : - \frac{2 ^{\frac{3}{4}}}{4} - i \frac{2 ^{\frac{3}{4}}}{4}

- \frac{1}{2 ^{\frac{5}{2}} \cdot \frac{1}{2 4 2}} - \frac{1}{2 4 2} i

乘

2^{\frac{5}{2}} \cdot \frac{1}{2 4 2} : 242

2^{\frac{5}{2}} \cdot \frac{1}{2 4 2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2 ^{\frac{5}{2}}}{2 4 2}

乘以:

1 \cdot 2^{\frac{5}{2}} = 2^{\frac{5}{2}}

= \frac{2 ^{\frac{5}{2}}}{2 4 2}

消掉

\frac{2 ^{\frac{5}{2}}}{2 4 2} : \frac{2 ^{\frac{3}{2}}}{4 2}

\frac{2 ^{\frac{5}{2}}}{2 4 2}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

\frac{2 ^{\frac{5}{2}}}{2} = 2^{\frac{5}{2} - 1}

= \frac{2 ^{\frac{5}{2} - 1}}{4 2}

数字相减:

\frac{5}{2} - 1 = \frac{3}{2}

= \frac{2 ^{\frac{3}{2}}}{4 2}

= \frac{2 ^{\frac{3}{2}}}{4 2}

消掉

\frac{2 ^{\frac{3}{2}}}{4 2} : 2^{\frac{5}{4}}

\frac{2 ^{\frac{3}{2}}}{4 2}

使用根式运算法则:

n a = a^{\frac{1}{n}}

42 = 2^{\frac{1}{4}}

= \frac{2 ^{\frac{3}{2}}}{2 ^{\frac{1}{4}}}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

\frac{2 ^{\frac{3}{2}}}{2 ^{\frac{1}{4}}} = 2^{\frac{3}{2} - \frac{1}{4}}

= 2^{\frac{3}{2} - \frac{1}{4}}

数字相减:

\frac{3}{2} - \frac{1}{4} = \frac{5}{4}

= 2^{\frac{5}{4}}

= 2^{\frac{5}{4}}

2^{\frac{5}{4}} = 242

2^{\frac{5}{4}}

2^{\frac{5}{4}} = 2^{1 + \frac{1}{4}}

= 2^{1 + \frac{1}{4}}

使用指数法则:

x^{a + b} = x^{a} x^{b}

= 2^{1} \cdot 2^{\frac{1}{4}}

整理后得

= 242

= 242

= - \frac{1}{2 4 2} - i \frac{1}{2 4 2}

乘

\frac{1}{2 4 2} i : \frac{i}{2 4 2}

\frac{1}{2 4 2} i

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 i}{2 4 2}

乘以:

1 i = i

= \frac{i}{2 4 2}

= - \frac{1}{2 4 2} - \frac{i}{2 4 2}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 1 - i}{2 4 2}

\frac{- 1 - i}{2 4 2}

有理化:

\frac{2 ^{\frac{3}{4}} ( - 1 - i )}{4}

\frac{- 1 - i}{2 4 2}

乘以共轭根式

\frac{2 ^{\frac{3}{4}}}{2 ^{\frac{3}{4}}}

= \frac{( - 1 - i ) \cdot 2 ^{\frac{3}{4}}}{2 4 2 \cdot 2 ^{\frac{3}{4}}}

242 \cdot 2^{\frac{3}{4}} = 4

242 \cdot 2^{\frac{3}{4}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2^{\frac{3}{4}} 42 = 2 \cdot 2^{\frac{3}{4}} \cdot 2^{\frac{1}{4}} = 2^{1 + \frac{3}{4} + \frac{1}{4}}

= 2^{1 + \frac{3}{4} + \frac{1}{4}}

化简

1 + \frac{3}{4} + \frac{1}{4} : 2

1 + \frac{3}{4} + \frac{1}{4}

将项转换为分式:

1 = \frac{1}{1}

= \frac{1}{1} + \frac{3}{4} + \frac{1}{4}

1, 4, 4

的最小公倍数:

4

1, 4, 4

最小公倍数 (LCM)

1

质因数分解

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

计算出由至少在以下一个数字中出现的因数组成的数字:

1, 4, 4

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

4

对于

\frac{1}{1} :

将分母和分子乘以

4

\frac{1}{1} = \frac{1 \cdot 4}{1 \cdot 4} = \frac{4}{4}

= \frac{4}{4} + \frac{3}{4} + \frac{1}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{4 + 3 + 1}{4}

数字相加:

4 + 3 + 1 = 8

= \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2 ^{\frac{3}{4}} ( - 1 - i )}{4}

= \frac{2 ^{\frac{3}{4}} ( - 1 - i )}{4}

将

\frac{2 ^{\frac{3}{4}} ( - 1 - i )}{4}

改写成标准复数形式:

- \frac{2 ^{\frac{3}{4}}}{4} - \frac{2 ^{\frac{3}{4}}}{4} i

\frac{2 ^{\frac{3}{4}} ( - 1 - i )}{4}

分解

4 : 2^{2}

因式分解

4 = 2^{2}

= \frac{2 ^{\frac{3}{4}} ( - 1 - i )}{2 ^{2}}

消掉

\frac{2 ^{\frac{3}{4}} ( - 1 - i )}{2 ^{2}} : \frac{- 1 - i}{2 ^{\frac{5}{4}}}

\frac{2 ^{\frac{3}{4}} ( - 1 - i )}{2 ^{2}}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{\frac{3}{4}}}{2 ^{2}} = \frac{1}{2 ^{2 - \frac{3}{4}}}

= \frac{- 1 - i}{2 ^{2 - \frac{3}{4}}}

数字相减:

2 - \frac{3}{4} = \frac{5}{4}

= \frac{- 1 - i}{2 ^{\frac{5}{4}}}

= \frac{- 1 - i}{2 ^{\frac{5}{4}}}

2^{\frac{5}{4}} = 242

2^{\frac{5}{4}}

2^{\frac{5}{4}} = 2^{1 + \frac{1}{4}}

= 2^{1 + \frac{1}{4}}

使用指数法则:

x^{a + b} = x^{a} x^{b}

= 2^{1} \cdot 2^{\frac{1}{4}}

整理后得

= 242

= \frac{- 1 - i}{2 4 2}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 - i}{2 4 2} = - \frac{1}{2 4 2} - \frac{i}{2 4 2}

= - \frac{1}{2 4 2} - \frac{i}{2 4 2}

- \frac{1}{2 4 2} = - \frac{2 ^{\frac{3}{4}}}{4}

- \frac{1}{2 4 2}

乘以共轭根式

\frac{2 ^{\frac{3}{4}}}{2 ^{\frac{3}{4}}}

= - \frac{1 \cdot 2 ^{\frac{3}{4}}}{2 4 2 \cdot 2 ^{\frac{3}{4}}}

1 \cdot 2^{\frac{3}{4}} = 2^{\frac{3}{4}}

242 \cdot 2^{\frac{3}{4}} = 4

242 \cdot 2^{\frac{3}{4}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2^{\frac{3}{4}} 42 = 2 \cdot 2^{\frac{3}{4}} \cdot 2^{\frac{1}{4}} = 2^{1 + \frac{3}{4} + \frac{1}{4}}

= 2^{1 + \frac{3}{4} + \frac{1}{4}}

化简

1 + \frac{3}{4} + \frac{1}{4} : 2

1 + \frac{3}{4} + \frac{1}{4}

将项转换为分式:

1 = \frac{1}{1}

= \frac{1}{1} + \frac{3}{4} + \frac{1}{4}

1, 4, 4

的最小公倍数:

4

1, 4, 4

最小公倍数 (LCM)

1

质因数分解

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

计算出由至少在以下一个数字中出现的因数组成的数字:

1, 4, 4

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

4

对于

\frac{1}{1} :

将分母和分子乘以

4

\frac{1}{1} = \frac{1 \cdot 4}{1 \cdot 4} = \frac{4}{4}

= \frac{4}{4} + \frac{3}{4} + \frac{1}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{4 + 3 + 1}{4}

数字相加:

4 + 3 + 1 = 8

= \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= 2

= 2^{2}

2^{2} = 4

= 4

= - \frac{2 ^{\frac{3}{4}}}{4}

= - \frac{1}{2 4 2} - \frac{2 ^{\frac{3}{4}}}{4} i

- \frac{1}{2 4 2} = - \frac{2 ^{\frac{3}{4}}}{4}

- \frac{1}{2 4 2}

乘以共轭根式

\frac{2 ^{\frac{3}{4}}}{2 ^{\frac{3}{4}}}

= - \frac{1 \cdot 2 ^{\frac{3}{4}}}{2 4 2 \cdot 2 ^{\frac{3}{4}}}

1 \cdot 2^{\frac{3}{4}} = 2^{\frac{3}{4}}

242 \cdot 2^{\frac{3}{4}} = 4

242 \cdot 2^{\frac{3}{4}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2^{\frac{3}{4}} 42 = 2 \cdot 2^{\frac{3}{4}} \cdot 2^{\frac{1}{4}} = 2^{1 + \frac{3}{4} + \frac{1}{4}}

= 2^{1 + \frac{3}{4} + \frac{1}{4}}

化简

1 + \frac{3}{4} + \frac{1}{4} : 2

1 + \frac{3}{4} + \frac{1}{4}

将项转换为分式:

1 = \frac{1}{1}

= \frac{1}{1} + \frac{3}{4} + \frac{1}{4}

1, 4, 4

的最小公倍数:

4

1, 4, 4

最小公倍数 (LCM)

1

质因数分解

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

计算出由至少在以下一个数字中出现的因数组成的数字:

1, 4, 4

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

4

对于

\frac{1}{1} :

将分母和分子乘以

4

\frac{1}{1} = \frac{1 \cdot 4}{1 \cdot 4} = \frac{4}{4}

= \frac{4}{4} + \frac{3}{4} + \frac{1}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{4 + 3 + 1}{4}

数字相加:

4 + 3 + 1 = 8

= \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= 2

= 2^{2}

2^{2} = 4

= 4

= - \frac{2 ^{\frac{3}{4}}}{4}

= - \frac{2 ^{\frac{3}{4}}}{4} - \frac{2 ^{\frac{3}{4}}}{4} i

= - \frac{2 ^{\frac{3}{4}}}{4} - \frac{2 ^{\frac{3}{4}}}{4} i

无解

sin (x) = - \frac{1}{2 ^{\frac{5}{4}}} + \frac{1}{2 4 2} i :

无解

sin (x) = - \frac{1}{2 ^{\frac{5}{4}}} + \frac{1}{2 4 2} i

化简

- \frac{1}{2 ^{\frac{5}{4}}} + \frac{1}{2 4 2} i : - \frac{2 ^{\frac{3}{4}}}{4} + i \frac{2 ^{\frac{3}{4}}}{4}

- \frac{1}{2 ^{\frac{5}{4}}} + \frac{1}{2 4 2} i

2^{\frac{5}{4}} = 242

2^{\frac{5}{4}}

2^{\frac{5}{4}} = 2^{1 + \frac{1}{4}}

= 2^{1 + \frac{1}{4}}

使用指数法则:

x^{a + b} = x^{a} x^{b}

= 2^{1} \cdot 2^{\frac{1}{4}}

整理后得

= 242

= - \frac{1}{2 4 2} + i \frac{1}{2 4 2}

乘

\frac{1}{2 4 2} i : \frac{i}{2 4 2}

\frac{1}{2 4 2} i

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 i}{2 4 2}

乘以:

1 i = i

= \frac{i}{2 4 2}

= - \frac{1}{2 4 2} + \frac{i}{2 4 2}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 1 + i}{2 4 2}

\frac{- 1 + i}{2 4 2}

有理化:

\frac{2 ^{\frac{3}{4}} ( - 1 + i )}{4}

\frac{- 1 + i}{2 4 2}

乘以共轭根式

\frac{2 ^{\frac{3}{4}}}{2 ^{\frac{3}{4}}}

= \frac{( - 1 + i ) \cdot 2 ^{\frac{3}{4}}}{2 4 2 \cdot 2 ^{\frac{3}{4}}}

242 \cdot 2^{\frac{3}{4}} = 4

242 \cdot 2^{\frac{3}{4}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2^{\frac{3}{4}} 42 = 2 \cdot 2^{\frac{3}{4}} \cdot 2^{\frac{1}{4}} = 2^{1 + \frac{3}{4} + \frac{1}{4}}

= 2^{1 + \frac{3}{4} + \frac{1}{4}}

化简

1 + \frac{3}{4} + \frac{1}{4} : 2

1 + \frac{3}{4} + \frac{1}{4}

将项转换为分式:

1 = \frac{1}{1}

= \frac{1}{1} + \frac{3}{4} + \frac{1}{4}

1, 4, 4

的最小公倍数:

4

1, 4, 4

最小公倍数 (LCM)

1

质因数分解

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

计算出由至少在以下一个数字中出现的因数组成的数字:

1, 4, 4

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

4

对于

\frac{1}{1} :

将分母和分子乘以

4

\frac{1}{1} = \frac{1 \cdot 4}{1 \cdot 4} = \frac{4}{4}

= \frac{4}{4} + \frac{3}{4} + \frac{1}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{4 + 3 + 1}{4}

数字相加:

4 + 3 + 1 = 8

= \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2 ^{\frac{3}{4}} ( - 1 + i )}{4}

= \frac{2 ^{\frac{3}{4}} ( - 1 + i )}{4}

将

\frac{2 ^{\frac{3}{4}} ( - 1 + i )}{4}

改写成标准复数形式:

- \frac{2 ^{\frac{3}{4}}}{4} + \frac{2 ^{\frac{3}{4}}}{4} i

\frac{2 ^{\frac{3}{4}} ( - 1 + i )}{4}

分解

4 : 2^{2}

因式分解

4 = 2^{2}

= \frac{2 ^{\frac{3}{4}} ( - 1 + i )}{2 ^{2}}

消掉

\frac{2 ^{\frac{3}{4}} ( - 1 + i )}{2 ^{2}} : \frac{- 1 + i}{2 ^{\frac{5}{4}}}

\frac{2 ^{\frac{3}{4}} ( - 1 + i )}{2 ^{2}}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{\frac{3}{4}}}{2 ^{2}} = \frac{1}{2 ^{2 - \frac{3}{4}}}

= \frac{- 1 + i}{2 ^{2 - \frac{3}{4}}}

数字相减:

2 - \frac{3}{4} = \frac{5}{4}

= \frac{- 1 + i}{2 ^{\frac{5}{4}}}

= \frac{- 1 + i}{2 ^{\frac{5}{4}}}

2^{\frac{5}{4}} = 242

2^{\frac{5}{4}}

2^{\frac{5}{4}} = 2^{1 + \frac{1}{4}}

= 2^{1 + \frac{1}{4}}

使用指数法则:

x^{a + b} = x^{a} x^{b}

= 2^{1} \cdot 2^{\frac{1}{4}}

整理后得

= 242

= \frac{- 1 + i}{2 4 2}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 + i}{2 4 2} = - \frac{1}{2 4 2} + \frac{i}{2 4 2}

= - \frac{1}{2 4 2} + \frac{i}{2 4 2}

\frac{1}{2 4 2} = \frac{2 ^{\frac{3}{4}}}{4}

\frac{1}{2 4 2}

乘以共轭根式

\frac{2 ^{\frac{3}{4}}}{2 ^{\frac{3}{4}}}

= \frac{1 \cdot 2 ^{\frac{3}{4}}}{2 4 2 \cdot 2 ^{\frac{3}{4}}}

1 \cdot 2^{\frac{3}{4}} = 2^{\frac{3}{4}}

242 \cdot 2^{\frac{3}{4}} = 4

242 \cdot 2^{\frac{3}{4}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2^{\frac{3}{4}} 42 = 2 \cdot 2^{\frac{3}{4}} \cdot 2^{\frac{1}{4}} = 2^{1 + \frac{3}{4} + \frac{1}{4}}

= 2^{1 + \frac{3}{4} + \frac{1}{4}}

化简

1 + \frac{3}{4} + \frac{1}{4} : 2

1 + \frac{3}{4} + \frac{1}{4}

将项转换为分式:

1 = \frac{1}{1}

= \frac{1}{1} + \frac{3}{4} + \frac{1}{4}

1, 4, 4

的最小公倍数:

4

1, 4, 4

最小公倍数 (LCM)

1

质因数分解

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

计算出由至少在以下一个数字中出现的因数组成的数字:

1, 4, 4

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

4

对于

\frac{1}{1} :

将分母和分子乘以

4

\frac{1}{1} = \frac{1 \cdot 4}{1 \cdot 4} = \frac{4}{4}

= \frac{4}{4} + \frac{3}{4} + \frac{1}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{4 + 3 + 1}{4}

数字相加:

4 + 3 + 1 = 8

= \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2 ^{\frac{3}{4}}}{4}

= - \frac{1}{2 4 2} + \frac{2 ^{\frac{3}{4}}}{4} i

- \frac{1}{2 4 2} = - \frac{2 ^{\frac{3}{4}}}{4}

- \frac{1}{2 4 2}

乘以共轭根式

\frac{2 ^{\frac{3}{4}}}{2 ^{\frac{3}{4}}}

= - \frac{1 \cdot 2 ^{\frac{3}{4}}}{2 4 2 \cdot 2 ^{\frac{3}{4}}}

1 \cdot 2^{\frac{3}{4}} = 2^{\frac{3}{4}}

242 \cdot 2^{\frac{3}{4}} = 4

242 \cdot 2^{\frac{3}{4}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2^{\frac{3}{4}} 42 = 2 \cdot 2^{\frac{3}{4}} \cdot 2^{\frac{1}{4}} = 2^{1 + \frac{3}{4} + \frac{1}{4}}

= 2^{1 + \frac{3}{4} + \frac{1}{4}}

化简

1 + \frac{3}{4} + \frac{1}{4} : 2

1 + \frac{3}{4} + \frac{1}{4}

将项转换为分式:

1 = \frac{1}{1}

= \frac{1}{1} + \frac{3}{4} + \frac{1}{4}

1, 4, 4

的最小公倍数:

4

1, 4, 4

最小公倍数 (LCM)

1

质因数分解

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

计算出由至少在以下一个数字中出现的因数组成的数字:

1, 4, 4

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

4

对于

\frac{1}{1} :

将分母和分子乘以

4

\frac{1}{1} = \frac{1 \cdot 4}{1 \cdot 4} = \frac{4}{4}

= \frac{4}{4} + \frac{3}{4} + \frac{1}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{4 + 3 + 1}{4}

数字相加:

4 + 3 + 1 = 8

= \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= 2

= 2^{2}

2^{2} = 4

= 4

= - \frac{2 ^{\frac{3}{4}}}{4}

= - \frac{2 ^{\frac{3}{4}}}{4} + \frac{2 ^{\frac{3}{4}}}{4} i

= - \frac{2 ^{\frac{3}{4}}}{4} + \frac{2 ^{\frac{3}{4}}}{4} i

无解

sin (x) = \frac{1}{2 ^{\frac{5}{4}}} - \frac{1}{2 4 2} i :

无解

sin (x) = \frac{1}{2 ^{\frac{5}{4}}} - \frac{1}{2 4 2} i

化简

\frac{1}{2 ^{\frac{5}{4}}} - \frac{1}{2 4 2} i : \frac{2 ^{\frac{3}{4}}}{4} - i \frac{2 ^{\frac{3}{4}}}{4}

\frac{1}{2 ^{\frac{5}{4}}} - \frac{1}{2 4 2} i

2^{\frac{5}{4}} = 242

2^{\frac{5}{4}}

2^{\frac{5}{4}} = 2^{1 + \frac{1}{4}}

= 2^{1 + \frac{1}{4}}

使用指数法则:

x^{a + b} = x^{a} x^{b}

= 2^{1} \cdot 2^{\frac{1}{4}}

整理后得

= 242

= \frac{1}{2 4 2} - i \frac{1}{2 4 2}

乘

\frac{1}{2 4 2} i : \frac{i}{2 4 2}

\frac{1}{2 4 2} i

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 i}{2 4 2}

乘以:

1 i = i

= \frac{i}{2 4 2}

= \frac{1}{2 4 2} - \frac{i}{2 4 2}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 - i}{2 4 2}

\frac{1 - i}{2 4 2}

有理化:

\frac{2 ^{\frac{3}{4}} ( 1 - i )}{4}

\frac{1 - i}{2 4 2}

乘以共轭根式

\frac{2 ^{\frac{3}{4}}}{2 ^{\frac{3}{4}}}

= \frac{( 1 - i ) \cdot 2 ^{\frac{3}{4}}}{2 4 2 \cdot 2 ^{\frac{3}{4}}}

242 \cdot 2^{\frac{3}{4}} = 4

242 \cdot 2^{\frac{3}{4}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2^{\frac{3}{4}} 42 = 2 \cdot 2^{\frac{3}{4}} \cdot 2^{\frac{1}{4}} = 2^{1 + \frac{3}{4} + \frac{1}{4}}

= 2^{1 + \frac{3}{4} + \frac{1}{4}}

化简

1 + \frac{3}{4} + \frac{1}{4} : 2

1 + \frac{3}{4} + \frac{1}{4}

将项转换为分式:

1 = \frac{1}{1}

= \frac{1}{1} + \frac{3}{4} + \frac{1}{4}

1, 4, 4

的最小公倍数:

4

1, 4, 4

最小公倍数 (LCM)

1

质因数分解

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

计算出由至少在以下一个数字中出现的因数组成的数字:

1, 4, 4

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

4

对于

\frac{1}{1} :

将分母和分子乘以

4

\frac{1}{1} = \frac{1 \cdot 4}{1 \cdot 4} = \frac{4}{4}

= \frac{4}{4} + \frac{3}{4} + \frac{1}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{4 + 3 + 1}{4}

数字相加:

4 + 3 + 1 = 8

= \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2 ^{\frac{3}{4}} ( 1 - i )}{4}

= \frac{2 ^{\frac{3}{4}} ( 1 - i )}{4}

将

\frac{2 ^{\frac{3}{4}} ( 1 - i )}{4}

改写成标准复数形式:

\frac{2 ^{\frac{3}{4}}}{4} - \frac{2 ^{\frac{3}{4}}}{4} i

\frac{2 ^{\frac{3}{4}} ( 1 - i )}{4}

分解

4 : 2^{2}

因式分解

4 = 2^{2}

= \frac{2 ^{\frac{3}{4}} ( 1 - i )}{2 ^{2}}

消掉

\frac{2 ^{\frac{3}{4}} ( 1 - i )}{2 ^{2}} : \frac{1 - i}{2 ^{\frac{5}{4}}}

\frac{2 ^{\frac{3}{4}} ( 1 - i )}{2 ^{2}}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{\frac{3}{4}}}{2 ^{2}} = \frac{1}{2 ^{2 - \frac{3}{4}}}

= \frac{1 - i}{2 ^{2 - \frac{3}{4}}}

数字相减:

2 - \frac{3}{4} = \frac{5}{4}

= \frac{1 - i}{2 ^{\frac{5}{4}}}

= \frac{1 - i}{2 ^{\frac{5}{4}}}

2^{\frac{5}{4}} = 242

2^{\frac{5}{4}}

2^{\frac{5}{4}} = 2^{1 + \frac{1}{4}}

= 2^{1 + \frac{1}{4}}

使用指数法则:

x^{a + b} = x^{a} x^{b}

= 2^{1} \cdot 2^{\frac{1}{4}}

整理后得

= 242

= \frac{1 - i}{2 4 2}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{1 - i}{2 4 2} = \frac{1}{2 4 2} - \frac{i}{2 4 2}

= \frac{1}{2 4 2} - \frac{i}{2 4 2}

- \frac{1}{2 4 2} = - \frac{2 ^{\frac{3}{4}}}{4}

- \frac{1}{2 4 2}

乘以共轭根式

\frac{2 ^{\frac{3}{4}}}{2 ^{\frac{3}{4}}}

= - \frac{1 \cdot 2 ^{\frac{3}{4}}}{2 4 2 \cdot 2 ^{\frac{3}{4}}}

1 \cdot 2^{\frac{3}{4}} = 2^{\frac{3}{4}}

242 \cdot 2^{\frac{3}{4}} = 4

242 \cdot 2^{\frac{3}{4}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2^{\frac{3}{4}} 42 = 2 \cdot 2^{\frac{3}{4}} \cdot 2^{\frac{1}{4}} = 2^{1 + \frac{3}{4} + \frac{1}{4}}

= 2^{1 + \frac{3}{4} + \frac{1}{4}}

化简

1 + \frac{3}{4} + \frac{1}{4} : 2

1 + \frac{3}{4} + \frac{1}{4}

将项转换为分式:

1 = \frac{1}{1}

= \frac{1}{1} + \frac{3}{4} + \frac{1}{4}

1, 4, 4

的最小公倍数:

4

1, 4, 4

最小公倍数 (LCM)

1

质因数分解

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

计算出由至少在以下一个数字中出现的因数组成的数字:

1, 4, 4

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

4

对于

\frac{1}{1} :

将分母和分子乘以

4

\frac{1}{1} = \frac{1 \cdot 4}{1 \cdot 4} = \frac{4}{4}

= \frac{4}{4} + \frac{3}{4} + \frac{1}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{4 + 3 + 1}{4}

数字相加:

4 + 3 + 1 = 8

= \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= 2

= 2^{2}

2^{2} = 4

= 4

= - \frac{2 ^{\frac{3}{4}}}{4}

= \frac{1}{2 4 2} - \frac{2 ^{\frac{3}{4}}}{4} i

\frac{1}{2 4 2} = \frac{2 ^{\frac{3}{4}}}{4}

\frac{1}{2 4 2}

乘以共轭根式

\frac{2 ^{\frac{3}{4}}}{2 ^{\frac{3}{4}}}

= \frac{1 \cdot 2 ^{\frac{3}{4}}}{2 4 2 \cdot 2 ^{\frac{3}{4}}}

1 \cdot 2^{\frac{3}{4}} = 2^{\frac{3}{4}}

242 \cdot 2^{\frac{3}{4}} = 4

242 \cdot 2^{\frac{3}{4}}

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2^{\frac{3}{4}} 42 = 2 \cdot 2^{\frac{3}{4}} \cdot 2^{\frac{1}{4}} = 2^{1 + \frac{3}{4} + \frac{1}{4}}

= 2^{1 + \frac{3}{4} + \frac{1}{4}}

化简

1 + \frac{3}{4} + \frac{1}{4} : 2

1 + \frac{3}{4} + \frac{1}{4}

将项转换为分式:

1 = \frac{1}{1}

= \frac{1}{1} + \frac{3}{4} + \frac{1}{4}

1, 4, 4

的最小公倍数:

4

1, 4, 4

最小公倍数 (LCM)

1

质因数分解

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

计算出由至少在以下一个数字中出现的因数组成的数字:

1, 4, 4

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

4

对于

\frac{1}{1} :

将分母和分子乘以

4

\frac{1}{1} = \frac{1 \cdot 4}{1 \cdot 4} = \frac{4}{4}

= \frac{4}{4} + \frac{3}{4} + \frac{1}{4}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{4 + 3 + 1}{4}

数字相加:

4 + 3 + 1 = 8

= \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2 ^{\frac{3}{4}}}{4}

= \frac{2 ^{\frac{3}{4}}}{4} - \frac{2 ^{\frac{3}{4}}}{4} i

= \frac{2 ^{\frac{3}{4}}}{4} - \frac{2 ^{\frac{3}{4}}}{4} i

无解

合并所有解

x \in R 无解

作图

流行的例子

sin(θ)=0.321

sin (θ) = 0.321

sin(x+75)=(sqrt(3))/2

sin (x + 7 5^{\circ}) = \frac{3}{2}

tan(x/6)+sqrt(3)=0

tan (\frac{x}{6}) + 3 = 0

solvefor k,6(-cos(k/2)+1)=1.5

so l v e f or k, 6 (- cos (\frac{k}{2}) + 1) = 1.5

tan(x)=sqrt(3),0<= x<2pi

tan (x) = 3, 0 \leq x < 2 π