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受欢迎的三角函数 >

sin^2(x)-csc^2(x)=tan^2(x)-cot^2(x)

解答

sin^{2} (x) - csc^{2} (x) = tan^{2} (x) - cot^{2} (x)

解答

x \in R 无解

求解步骤

sin^{2} (x) - csc^{2} (x) = tan^{2} (x) - cot^{2} (x)

两边减去

tan^{2} (x) - cot^{2} (x)

sin^{2} (x) - csc^{2} (x) - tan^{2} (x) + cot^{2} (x) = 0

用 sin, cos 表示

cot^{2} (x) - csc^{2} (x) + sin^{2} (x) - tan^{2} (x)

使用基本三角恒等式:

cot (x) = \frac{cos ( x )}{sin ( x )}

= (\frac{cos ( x )}{sin ( x )})^{2} - csc^{2} (x) + sin^{2} (x) - tan^{2} (x)

使用基本三角恒等式:

csc (x) = \frac{1}{sin ( x )}

= (\frac{cos ( x )}{sin ( x )})^{2} - (\frac{1}{sin ( x )})^{2} + sin^{2} (x) - tan^{2} (x)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= (\frac{cos ( x )}{sin ( x )})^{2} - (\frac{1}{sin ( x )})^{2} + sin^{2} (x) - (\frac{sin ( x )}{cos ( x )})^{2}

化简

(\frac{cos ( x )}{sin ( x )})^{2} - (\frac{1}{sin ( x )})^{2} + sin^{2} (x) - (\frac{sin ( x )}{cos ( x )})^{2} : \frac{cos ^{2} ( x ) ( cos ^{2} ( x ) - 1 ) + sin ^{4} ( x ) cos ^{2} ( x ) - sin ^{4} ( x )}{sin ^{2} ( x ) cos ^{2} ( x )}

(\frac{cos ( x )}{sin ( x )})^{2} - (\frac{1}{sin ( x )})^{2} + sin^{2} (x) - (\frac{sin ( x )}{cos ( x )})^{2}

(\frac{cos ( x )}{sin ( x )})^{2} = \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

(\frac{cos ( x )}{sin ( x )})^{2}

使用指数法则:

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

= \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

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

(\frac{1}{sin ( x )})^{2}

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

(\frac{sin ( x )}{cos ( x )})^{2} = \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

(\frac{sin ( x )}{cos ( x )})^{2}

使用指数法则:

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

= \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

= \frac{cos ^{2} ( x )}{sin ^{2} ( x )} - \frac{1}{sin ^{2} ( x )} + sin^{2} (x) - \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

合并分式

\frac{cos ^{2} ( x )}{sin ^{2} ( x )} - \frac{1}{sin ^{2} ( x )} : \frac{cos ^{2} ( x ) - 1}{sin ^{2} ( x )}

使用法则

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

= \frac{cos ^{2} ( x ) - 1}{sin ^{2} ( x )}

= \frac{cos ^{2} ( x ) - 1}{sin ^{2} ( x )} + sin^{2} (x) - \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

将项转换为分式:

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

= \frac{cos ^{2} ( x ) - 1}{sin ^{2} ( x )} + \frac{sin ^{2} ( x )}{1} - \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

sin^{2} (x), 1, cos^{2} (x)

的最小公倍数:

sin^{2} (x) cos^{2} (x)

sin^{2} (x), 1, cos^{2} (x)

最小公倍数 (LCM)

计算出由至少在以下一个因式表达式中出现的因子组成的表达式

= sin^{2} (x) cos^{2} (x)

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

sin^{2} (x) cos^{2} (x)

对于

\frac{cos ^{2} ( x ) - 1}{sin ^{2} ( x )} :

将分母和分子乘以

cos^{2} (x)

\frac{cos ^{2} ( x ) - 1}{sin ^{2} ( x )} = \frac{( cos ^{2} ( x ) - 1 ) cos ^{2} ( x )}{sin ^{2} ( x ) cos ^{2} ( x )}

对于

\frac{sin ^{2} ( x )}{1} :

将分母和分子乘以

sin^{2} (x) cos^{2} (x)

\frac{sin ^{2} ( x )}{1} = \frac{sin ^{2} ( x ) sin ^{2} ( x ) cos ^{2} ( x )}{1 \cdot sin ^{2} ( x ) cos ^{2} ( x )} = \frac{sin ^{4} ( x ) cos ^{2} ( x )}{sin ^{2} ( x ) cos ^{2} ( x )}

对于

\frac{sin ^{2} ( x )}{cos ^{2} ( x )} :

将分母和分子乘以

sin^{2} (x)

\frac{sin ^{2} ( x )}{cos ^{2} ( x )} = \frac{sin ^{2} ( x ) sin ^{2} ( x )}{cos ^{2} ( x ) sin ^{2} ( x )} = \frac{sin ^{4} ( x )}{sin ^{2} ( x ) cos ^{2} ( x )}

= \frac{( cos ^{2} ( x ) - 1 ) cos ^{2} ( x )}{sin ^{2} ( x ) cos ^{2} ( x )} + \frac{sin ^{4} ( x ) cos ^{2} ( x )}{sin ^{2} ( x ) cos ^{2} ( x )} - \frac{sin ^{4} ( x )}{sin ^{2} ( x ) cos ^{2} ( x )}

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{( cos ^{2} ( x ) - 1 ) cos ^{2} ( x ) + sin ^{4} ( x ) cos ^{2} ( x ) - sin ^{4} ( x )}{sin ^{2} ( x ) cos ^{2} ( x )}

= \frac{cos ^{2} ( x ) ( cos ^{2} ( x ) - 1 ) + sin ^{4} ( x ) cos ^{2} ( x ) - sin ^{4} ( x )}{sin ^{2} ( x ) cos ^{2} ( x )}

\frac{- sin ^{4} ( x ) + ( - 1 + cos ^{2} ( x ) ) cos ^{2} ( x ) + cos ^{2} ( x ) sin ^{4} ( x )}{cos ^{2} ( x ) sin ^{2} ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- sin^{4} (x) + (- 1 + cos^{2} (x)) cos^{2} (x) + cos^{2} (x) sin^{4} (x) = 0

分解

- sin^{4} (x) + (- 1 + cos^{2} (x)) cos^{2} (x) + cos^{2} (x) sin^{4} (x) : (- 1 + cos (x)) (1 + cos (x)) (sin^{4} (x) + cos^{2} (x))

- sin^{4} (x) + (- 1 + cos^{2} (x)) cos^{2} (x) + cos^{2} (x) sin^{4} (x)

分解

- 1 + cos^{2} (x) : (cos (x) + 1) (cos (x) - 1)

- 1 + cos^{2} (x)

将

1

改写为

1^{2}

= cos^{2} (x) - 1^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

cos^{2} (x) - 1^{2} = (cos (x) + 1) (cos (x) - 1)

= (cos (x) + 1) (cos (x) - 1)

= - sin^{4} (x) + cos^{2} (x) (cos (x) + 1) (cos (x) - 1) + sin^{4} (x) cos^{2} (x)

因式分解出通项

sin^{4} (x)

= sin^{4} (x) (- 1 + cos^{2} (x)) + (1 + cos (x)) (- 1 + cos (x)) cos^{2} (x)

分解

cos^{2} (x) - 1 : (cos (x) + 1) (cos (x) - 1)

- 1 + cos^{2} (x)

将

1

改写为

1^{2}

= cos^{2} (x) - 1^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

cos^{2} (x) - 1^{2} = (cos (x) + 1) (cos (x) - 1)

= (cos (x) + 1) (cos (x) - 1)

= sin^{4} (x) (cos (x) + 1) (cos (x) - 1) + cos^{2} (x) (cos (x) + 1) (cos (x) - 1)

因式分解出通项

(- 1 + cos (x)) (1 + cos (x))

= (- 1 + cos (x)) (1 + cos (x)) (sin^{4} (x) + cos^{2} (x))

(- 1 + cos (x)) (1 + cos (x)) (sin^{4} (x) + cos^{2} (x)) = 0

分别求解每个部分

- 1 + cos (x) = 0 or 1 + cos (x) = 0 or sin^{4} (x) + cos^{2} (x) = 0

- 1 + cos (x) = 0 : x = 2 πn

- 1 + cos (x) = 0

将

1

到右边

- 1 + cos (x) = 0

两边加上

1

- 1 + cos (x) + 1 = 0 + 1

化简

cos (x) = 1

cos (x) = 1

cos (x) = 1

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

1 + cos (x) = 0 : x = π + 2 πn

1 + cos (x) = 0

将

1

到右边

1 + cos (x) = 0

两边减去

1

1 + cos (x) - 1 = 0 - 1

化简

cos (x) = - 1

cos (x) = - 1

cos (x) = - 1

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

sin^{4} (x) + cos^{2} (x) = 0 :

无解

sin^{4} (x) + cos^{2} (x) = 0

使用三角恒等式改写

cos^{2} (x) + sin^{4} (x)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= 1 - sin^{2} (x) + sin^{4} (x)

1 - sin^{2} (x) + sin^{4} (x) = 0

用替代法求解

1 - sin^{2} (x) + sin^{4} (x) = 0

令

: sin (x) = u

1 - u^{2} + u^{4} = 0

1 - u^{2} + u^{4} = 0 : u = \frac{3}{2} + \frac{1}{2} i, u = - \frac{3}{2} - \frac{1}{2} i, u = - \frac{3}{2} + \frac{1}{2} i, u = \frac{3}{2} - \frac{1}{2} i

1 - u^{2} + u^{4} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{4} - u^{2} + 1 = 0

用

a = u^{2}

和

a^{2} = u^{4}

改写方程式

a^{2} - a + 1 = 0

解

a^{2} - a + 1 = 0 : a = \frac{1}{2} + i \frac{3}{2}, a = \frac{1}{2} - i \frac{3}{2}

a^{2} - a + 1 = 0

使用求根公式求解

a^{2} - a + 1 = 0

二次方程求根公式:

若

a = 1, b = - 1, c = 1

a_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

a_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

化简

(- 1)^{2} - 4 \cdot 1 \cdot 1 : 3 i

(- 1)^{2} - 4 \cdot 1 \cdot 1

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 - 4

数字相减:

1 - 4 = - 3

= - 3

使用根式运算法则:

- a = - 1 a

- 3 = - 1 3

= - 1 3

使用虚数运算法则:

- 1 = i

= 3 i

a_{1, 2} = \frac{- ( - 1 ) \pm 3 i}{2 \cdot 1}

将解分隔开

a_{1} = \frac{- ( - 1 ) + 3 i}{2 \cdot 1}, a_{2} = \frac{- ( - 1 ) - 3 i}{2 \cdot 1}

a = \frac{- ( - 1 ) + 3 i}{2 \cdot 1} : \frac{1}{2} + i \frac{3}{2}

\frac{- ( - 1 ) + 3 i}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 + 3 i}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 + 3 i}{2}

将

\frac{1 + 3 i}{2}

改写成标准复数形式:

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

\frac{1 + 3 i}{2}

使用分式法则:

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

\frac{1 + 3 i}{2} = \frac{1}{2} + \frac{3 i}{2}

= \frac{1}{2} + \frac{3 i}{2}

= \frac{1}{2} + \frac{3}{2} i

a = \frac{- ( - 1 ) - 3 i}{2 \cdot 1} : \frac{1}{2} - i \frac{3}{2}

\frac{- ( - 1 ) - 3 i}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 - 3 i}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 - 3 i}{2}

将

\frac{1 - 3 i}{2}

改写成标准复数形式:

\frac{1}{2} - \frac{3}{2} i

\frac{1 - 3 i}{2}

使用分式法则:

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

\frac{1 - 3 i}{2} = \frac{1}{2} - \frac{3 i}{2}

= \frac{1}{2} - \frac{3 i}{2}

= \frac{1}{2} - \frac{3}{2} i

二次方程组的解是:

a = \frac{1}{2} + i \frac{3}{2}, a = \frac{1}{2} - i \frac{3}{2}

a = \frac{1}{2} + i \frac{3}{2}, a = \frac{1}{2} - i \frac{3}{2}

代回

a = u^{2}

，求解

u

解

u^{2} = \frac{1}{2} + i \frac{3}{2} : u = \frac{3}{2} + \frac{1}{2} i, u = - \frac{3}{2} - \frac{1}{2} i

u^{2} = \frac{1}{2} + i \frac{3}{2}

替代

u = a + bi

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

展开

(a + bi)^{2} : (a^{2} - b^{2}) + 2 iab

(a + bi)^{2}

= (a + ib)^{2}

使用完全平方公式:

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

a = a, b = bi

= a^{2} + 2 abi + (bi)^{2}

(bi)^{2} = - b^{2}

(bi)^{2}

使用指数法则:

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

= i^{2} b^{2}

i^{2} = - 1

i^{2}

使用虚数运算法则:

i^{2} = - 1

= - 1

= (- 1) b^{2}

整理后得

= - b^{2}

= a^{2} + 2 iab - b^{2}

将

a^{2} + 2 iab - b^{2}

改写成标准复数形式:

(a^{2} - b^{2}) + 2 abi

a^{2} + 2 iab - b^{2}

将复数的实部和虚部分组

= (a^{2} - b^{2}) + 2 abi

= (a^{2} - b^{2}) + 2 abi

(a^{2} - b^{2}) + 2 iab = \frac{1}{2} + i \frac{3}{2}

复数仅在实部和虚部均相等时才相等改写为方程组:

[a^{2} - b^{2} = \frac{1}{2} 2 ab = \frac{3}{2}]

[a^{2} - b^{2} = \frac{1}{2} 2 ab = \frac{3}{2}] : (a = \frac{3}{2}, a = - \frac{3}{2}, b = \frac{1}{2} b = - \frac{1}{2})

[a^{2} - b^{2} = \frac{1}{2} 2 ab = \frac{3}{2}]

对于

2 ab = \frac{3}{2}

将

a

移到一边:

a = \frac{3}{4 b}

2 ab = \frac{3}{2}

两边除以

2 b

2 ab = \frac{3}{2}

两边除以

2 b

\frac{2 ab}{2 b} = \frac{\frac{3}{2}}{2 b}

化简

\frac{2 ab}{2 b} = \frac{\frac{3}{2}}{2 b}

化简

\frac{2 ab}{2 b} : a

\frac{2 ab}{2 b}

数字相除:

\frac{2}{2} = 1

= \frac{ab}{b}

约分:

b

= a

化简

\frac{\frac{3}{2}}{2 b} : \frac{3}{4 b}

\frac{\frac{3}{2}}{2 b}

使用分式法则:

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

= \frac{3}{2 \cdot 2 b}

数字相乘:

2 \cdot 2 = 4

= \frac{3}{4 b}

a = \frac{3}{4 b}

a = \frac{3}{4 b}

a = \frac{3}{4 b}

将解

a = \frac{3}{4 b}

代入

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

对于

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

，用

\frac{3}{4 b}

替代

a : b = \frac{1}{2}, b = - \frac{1}{2}

对于

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

，用

\frac{3}{4 b}

替代

a

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

解

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

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

乘以最小公倍数

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

化简

(\frac{3}{4 b})^{2} : \frac{3}{16 b ^{2}}

(\frac{3}{4 b})^{2}

使用指数法则:

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

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

使用指数法则:

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

(4 b)^{2} = 4^{2} b^{2}

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

(3)^{2} : 3

使用根式运算法则:

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

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

使用指数法则:

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

= 3^{\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

= 3

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

4^{2} = 16

= \frac{3}{16 b ^{2}}

\frac{3}{16 b ^{2}} - b^{2} = \frac{1}{2}

找到

16 b^{2}, 2

的最小公倍数:

16 b^{2}

16 b^{2}, 2

最小公倍数 (LCM)

16, 2

的最小公倍数:

16

16, 2

最小公倍数 (LCM)

16

质因数分解:

2 \cdot 2 \cdot 2 \cdot 2

16

16

除以

216 = 8 \cdot 2

= 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 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

16

或

2

中出现的最多次数

= 2 \cdot 2 \cdot 2 \cdot 2

数字相乘:

2 \cdot 2 \cdot 2 \cdot 2 = 16

= 16

计算出由出现在

16 b^{2}

或

2

中的因子组成的表达式

= 16 b^{2}

乘以最小公倍数=

16 b^{2}

\frac{3}{16 b ^{2}} \cdot 16 b^{2} - b^{2} \cdot 16 b^{2} = \frac{1}{2} \cdot 16 b^{2}

化简

\frac{3}{16 b ^{2}} \cdot 16 b^{2} - b^{2} \cdot 16 b^{2} = \frac{1}{2} \cdot 16 b^{2}

化简

\frac{3}{16 b ^{2}} \cdot 16 b^{2} : 3

\frac{3}{16 b ^{2}} \cdot 16 b^{2}

分式相乘:

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

= \frac{3 \cdot 16 b ^{2}}{16 b ^{2}}

约分:

16

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

约分:

b^{2}

= 3

化简

- b^{2} \cdot 16 b^{2} : - 16 b^{4}

- b^{2} \cdot 16 b^{2}

使用指数法则:

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

b^{2} b^{2} = b^{2 + 2}

= - 16 b^{2 + 2}

数字相加:

2 + 2 = 4

= - 16 b^{4}

化简

\frac{1}{2} \cdot 16 b^{2} : 8 b^{2}

\frac{1}{2} \cdot 16 b^{2}

分式相乘:

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

= \frac{1 \cdot 16}{2} b^{2}

\frac{1 \cdot 16}{2} = 8

\frac{1 \cdot 16}{2}

数字相乘:

1 \cdot 16 = 16

= \frac{16}{2}

数字相除:

\frac{16}{2} = 8

= 8

= 8 b^{2}

3 - 16 b^{4} = 8 b^{2}

3 - 16 b^{4} = 8 b^{2}

3 - 16 b^{4} = 8 b^{2}

解

3 - 16 b^{4} = 8 b^{2} : b = \frac{1}{2}, b = - \frac{1}{2}

3 - 16 b^{4} = 8 b^{2}

将

8 b^{2}

para o lado esquerdo

3 - 16 b^{4} = 8 b^{2}

两边减去

8 b^{2}

3 - 16 b^{4} - 8 b^{2} = 8 b^{2} - 8 b^{2}

化简

3 - 16 b^{4} - 8 b^{2} = 0

3 - 16 b^{4} - 8 b^{2} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- 16 b^{4} - 8 b^{2} + 3 = 0

用

u = b^{2}

和

u^{2} = b^{4}

改写方程式

- 16 u^{2} - 8 u + 3 = 0

解

- 16 u^{2} - 8 u + 3 = 0 : u = - \frac{3}{4}, u = \frac{1}{4}

- 16 u^{2} - 8 u + 3 = 0

使用求根公式求解

- 16 u^{2} - 8 u + 3 = 0

二次方程求根公式:

若

a = - 16, b = - 8, c = 3

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 16 ) \cdot 3}{2 ( - 16 )}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 16 ) \cdot 3}{2 ( - 16 )}

(- 8)^{2} - 4 (- 16) \cdot 3 = 16

(- 8)^{2} - 4 (- 16) \cdot 3

使用法则

- (- a) = a

= (- 8)^{2} + 4 \cdot 16 \cdot 3

使用指数法则:

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

若

n

是偶数

(- 8)^{2} = 8^{2}

= 8^{2} + 4 \cdot 16 \cdot 3

数字相乘:

4 \cdot 16 \cdot 3 = 192

= 8^{2} + 192

8^{2} = 64

= 64 + 192

数字相加:

64 + 192 = 256

= 256

因式分解数字:

256 = 1 6^{2}

= 1 6^{2}

使用根式运算法则:

n a^{n} = a

1 6^{2} = 16

= 16

u_{1, 2} = \frac{- ( - 8 ) \pm 16}{2 ( - 16 )}

将解分隔开

u_{1} = \frac{- ( - 8 ) + 16}{2 ( - 16 )}, u_{2} = \frac{- ( - 8 ) - 16}{2 ( - 16 )}

u = \frac{- ( - 8 ) + 16}{2 ( - 16 )} : - \frac{3}{4}

\frac{- ( - 8 ) + 16}{2 ( - 16 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{8 + 16}{- 2 \cdot 16}

数字相加:

8 + 16 = 24

= \frac{24}{- 2 \cdot 16}

数字相乘:

2 \cdot 16 = 32

= \frac{24}{- 32}

使用分式法则:

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

= - \frac{24}{32}

约分:

8

= - \frac{3}{4}

u = \frac{- ( - 8 ) - 16}{2 ( - 16 )} : \frac{1}{4}

\frac{- ( - 8 ) - 16}{2 ( - 16 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{8 - 16}{- 2 \cdot 16}

数字相减:

8 - 16 = - 8

= \frac{- 8}{- 2 \cdot 16}

数字相乘:

2 \cdot 16 = 32

= \frac{- 8}{- 32}

使用分式法则:

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

= \frac{8}{32}

约分:

8

= \frac{1}{4}

二次方程组的解是:

u = - \frac{3}{4}, u = \frac{1}{4}

u = - \frac{3}{4}, u = \frac{1}{4}

代回

u = b^{2}

，求解

b

解

b^{2} = - \frac{3}{4} : b \in R

无解

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

x^{2}

在

x

内不能为负

\in R

b \in R 无解

解

b^{2} = \frac{1}{4} : b = \frac{1}{2}, b = - \frac{1}{2}

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

对于

x^{2} = f (a)

解为

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

b = \frac{1}{4}, b = - \frac{1}{4}

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

\frac{1}{4}

使用根式运算法则:

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

= \frac{1}{4}

使用根式运算法则:

1 = 1

1 = 1

= \frac{1}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

a^{2} = a, a \geq 0

2^{2} = 2

= 2

= \frac{1}{2}

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

- \frac{1}{4}

使用根式运算法则:

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

= - \frac{1}{4}

使用根式运算法则:

1 = 1

1 = 1

= - \frac{1}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

a^{2} = a, a \geq 0

2^{2} = 2

= 2

= - \frac{1}{2}

b = \frac{1}{2}, b = - \frac{1}{2}

解为

b = \frac{1}{2}, b = - \frac{1}{2}

b = \frac{1}{2}, b = - \frac{1}{2}

验证解

找到无定义的点（奇点）:

b = 0

取

(\frac{3}{4 b})^{2} - b^{2}

的分母，令其等于零

解

4 b = 0 : b = 0

4 b = 0

两边除以

4

4 b = 0

两边除以

4

\frac{4 b}{4} = \frac{0}{4}

化简

b = 0

b = 0

以下点无定义

b = 0

将不在定义域的点与解相综合:

b = \frac{1}{2}, b = - \frac{1}{2}

将解

b = \frac{1}{2}, b = - \frac{1}{2}

代入

2 ab = \frac{3}{2}

对于

2 ab = \frac{3}{2}

，用

\frac{1}{2}

替代

b : a = \frac{3}{2}

对于

2 ab = \frac{3}{2}

，用

\frac{1}{2}

替代

b

2 a \frac{1}{2} = \frac{3}{2}

解

2 a \frac{1}{2} = \frac{3}{2} : a = \frac{3}{2}

2 a \frac{1}{2} = \frac{3}{2}

分式相乘:

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

\frac{1 \cdot 2}{2} a = \frac{3}{2}

约分:

2

a \cdot 1 = \frac{3}{2}

乘以:

a \cdot 1 = a

a = \frac{3}{2}

对于

2 ab = \frac{3}{2}

，用

- \frac{1}{2}

替代

b : a = - \frac{3}{2}

对于

2 ab = \frac{3}{2}

，用

- \frac{1}{2}

替代

b

2 a (- \frac{1}{2}) = \frac{3}{2}

解

2 a (- \frac{1}{2}) = \frac{3}{2} : a = - \frac{3}{2}

2 a (- \frac{1}{2}) = \frac{3}{2}

两边除以

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

2 a (- \frac{1}{2}) = \frac{3}{2}

两边除以

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

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} = \frac{\frac{3}{2}}{2 ( - \frac{1}{2} )}

化简

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} = \frac{\frac{3}{2}}{2 ( - \frac{1}{2} )}

化简

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} : a

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

去除括号:

(- a) = - a

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

使用分式法则:

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

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

乘

2 a \frac{1}{2} : a

2 a \frac{1}{2}

分式相乘:

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

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

约分:

2

= 1 \cdot a

乘以:

1 \cdot a = a

= a

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

乘

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

= \frac{a}{1}

使用法则

\frac{a}{1} = a

= a

化简

\frac{\frac{3}{2}}{2 ( - \frac{1}{2} )} : - \frac{3}{2}

\frac{\frac{3}{2}}{2 ( - \frac{1}{2} )}

去除括号:

(- a) = - a

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

使用分式法则:

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

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

使用分式法则:

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

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

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

数字相乘:

2 \cdot 2 = 4

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

乘

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

4 \cdot \frac{1}{2}

分式相乘:

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

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

数字相乘:

1 \cdot 4 = 4

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

= - \frac{3}{2}

a = - \frac{3}{2}

a = - \frac{3}{2}

a = - \frac{3}{2}

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

a = - \frac{3}{2}, b = - \frac{1}{2}

的解:

真

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

代入

a = - \frac{3}{2}, b = - \frac{1}{2}

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

整理后得

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

真

检验

a = \frac{3}{2}, b = \frac{1}{2}

的解:

真

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

代入

a = \frac{3}{2}, b = \frac{1}{2}

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

整理后得

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

真

将它们代入

2 ab = \frac{3}{2}

检验解是否符合

去除与方程不符的解。

检验

a = - \frac{3}{2}, b = - \frac{1}{2}

的解:

真

2 ab = \frac{3}{2}

代入

a = - \frac{3}{2}, b = - \frac{1}{2}

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

整理后得

\frac{3}{2} = \frac{3}{2}

真

检验

a = \frac{3}{2}, b = \frac{1}{2}

的解:

真

2 ab = \frac{3}{2}

代入

a = \frac{3}{2}, b = \frac{1}{2}

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

整理后得

\frac{3}{2} = \frac{3}{2}

真

因而，

a^{2} - b^{2} = \frac{1}{2}, 2 ab = \frac{3}{2}

最后的解是

(a = \frac{3}{2}, a = - \frac{3}{2}, b = \frac{1}{2} b = - \frac{1}{2})

u = a + bi

代回

u = \frac{3}{2} + \frac{1}{2} i, u = - \frac{3}{2} - \frac{1}{2} i

解

u^{2} = \frac{1}{2} - i \frac{3}{2} : u = - \frac{3}{2} + \frac{1}{2} i, u = \frac{3}{2} - \frac{1}{2} i

u^{2} = \frac{1}{2} - i \frac{3}{2}

替代

u = a + bi

(a + bi)^{2} = \frac{1}{2} - i \frac{3}{2}

展开

(a + bi)^{2} : (a^{2} - b^{2}) + 2 iab

(a + bi)^{2}

= (a + ib)^{2}

使用完全平方公式:

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

a = a, b = bi

= a^{2} + 2 abi + (bi)^{2}

(bi)^{2} = - b^{2}

(bi)^{2}

使用指数法则:

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

= i^{2} b^{2}

i^{2} = - 1

i^{2}

使用虚数运算法则:

i^{2} = - 1

= - 1

= (- 1) b^{2}

整理后得

= - b^{2}

= a^{2} + 2 iab - b^{2}

将

a^{2} + 2 iab - b^{2}

改写成标准复数形式:

(a^{2} - b^{2}) + 2 abi

a^{2} + 2 iab - b^{2}

将复数的实部和虚部分组

= (a^{2} - b^{2}) + 2 abi

= (a^{2} - b^{2}) + 2 abi

(a^{2} - b^{2}) + 2 iab = \frac{1}{2} - i \frac{3}{2}

复数仅在实部和虚部均相等时才相等改写为方程组:

[a^{2} - b^{2} = \frac{1}{2} 2 ab = - \frac{3}{2}]

[a^{2} - b^{2} = \frac{1}{2} 2 ab = - \frac{3}{2}] : (a = - \frac{3}{2}, a = \frac{3}{2}, b = \frac{1}{2} b = - \frac{1}{2})

[a^{2} - b^{2} = \frac{1}{2} 2 ab = - \frac{3}{2}]

对于

2 ab = - \frac{3}{2}

将

a

移到一边:

a = - \frac{3}{4 b}

2 ab = - \frac{3}{2}

两边除以

2 b

2 ab = - \frac{3}{2}

两边除以

2 b

\frac{2 ab}{2 b} = \frac{- \frac{3}{2}}{2 b}

化简

\frac{2 ab}{2 b} = \frac{- \frac{3}{2}}{2 b}

化简

\frac{2 ab}{2 b} : a

\frac{2 ab}{2 b}

数字相除:

\frac{2}{2} = 1

= \frac{ab}{b}

约分:

b

= a

化简

\frac{- \frac{3}{2}}{2 b} : - \frac{3}{4 b}

\frac{- \frac{3}{2}}{2 b}

使用分式法则:

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

= - \frac{\frac{3}{2}}{2 b}

使用分式法则:

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

\frac{\frac{3}{2}}{2 b} = \frac{3}{2 \cdot 2 b}

= - \frac{3}{2 \cdot 2 b}

数字相乘:

2 \cdot 2 = 4

= - \frac{3}{4 b}

a = - \frac{3}{4 b}

a = - \frac{3}{4 b}

a = - \frac{3}{4 b}

将解

a = - \frac{3}{4 b}

代入

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

对于

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

，用

- \frac{3}{4 b}

替代

a : b = \frac{1}{2}, b = - \frac{1}{2}

对于

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

，用

- \frac{3}{4 b}

替代

a

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

解

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

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

乘以最小公倍数

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

化简

(- \frac{3}{4 b})^{2} : \frac{3}{16 b ^{2}}

(- \frac{3}{4 b})^{2}

使用指数法则:

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

若

n

是偶数

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

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

使用指数法则:

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

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

使用指数法则:

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

(4 b)^{2} = 4^{2} b^{2}

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

(3)^{2} : 3

使用根式运算法则:

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

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

使用指数法则:

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

= 3^{\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

= 3

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

4^{2} = 16

= \frac{3}{16 b ^{2}}

\frac{3}{16 b ^{2}} - b^{2} = \frac{1}{2}

找到

16 b^{2}, 2

的最小公倍数:

16 b^{2}

16 b^{2}, 2

最小公倍数 (LCM)

16, 2

的最小公倍数:

16

16, 2

最小公倍数 (LCM)

16

质因数分解:

2 \cdot 2 \cdot 2 \cdot 2

16

16

除以

216 = 8 \cdot 2

= 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 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

16

或

2

中出现的最多次数

= 2 \cdot 2 \cdot 2 \cdot 2

数字相乘:

2 \cdot 2 \cdot 2 \cdot 2 = 16

= 16

计算出由出现在

16 b^{2}

或

2

中的因子组成的表达式

= 16 b^{2}

乘以最小公倍数=

16 b^{2}

\frac{3}{16 b ^{2}} \cdot 16 b^{2} - b^{2} \cdot 16 b^{2} = \frac{1}{2} \cdot 16 b^{2}

化简

\frac{3}{16 b ^{2}} \cdot 16 b^{2} - b^{2} \cdot 16 b^{2} = \frac{1}{2} \cdot 16 b^{2}

化简

\frac{3}{16 b ^{2}} \cdot 16 b^{2} : 3

\frac{3}{16 b ^{2}} \cdot 16 b^{2}

分式相乘:

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

= \frac{3 \cdot 16 b ^{2}}{16 b ^{2}}

约分:

16

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

约分:

b^{2}

= 3

化简

- b^{2} \cdot 16 b^{2} : - 16 b^{4}

- b^{2} \cdot 16 b^{2}

使用指数法则:

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

b^{2} b^{2} = b^{2 + 2}

= - 16 b^{2 + 2}

数字相加:

2 + 2 = 4

= - 16 b^{4}

化简

\frac{1}{2} \cdot 16 b^{2} : 8 b^{2}

\frac{1}{2} \cdot 16 b^{2}

分式相乘:

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

= \frac{1 \cdot 16}{2} b^{2}

\frac{1 \cdot 16}{2} = 8

\frac{1 \cdot 16}{2}

数字相乘:

1 \cdot 16 = 16

= \frac{16}{2}

数字相除:

\frac{16}{2} = 8

= 8

= 8 b^{2}

3 - 16 b^{4} = 8 b^{2}

3 - 16 b^{4} = 8 b^{2}

3 - 16 b^{4} = 8 b^{2}

解

3 - 16 b^{4} = 8 b^{2} : b = \frac{1}{2}, b = - \frac{1}{2}

3 - 16 b^{4} = 8 b^{2}

将

8 b^{2}

para o lado esquerdo

3 - 16 b^{4} = 8 b^{2}

两边减去

8 b^{2}

3 - 16 b^{4} - 8 b^{2} = 8 b^{2} - 8 b^{2}

化简

3 - 16 b^{4} - 8 b^{2} = 0

3 - 16 b^{4} - 8 b^{2} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- 16 b^{4} - 8 b^{2} + 3 = 0

用

u = b^{2}

和

u^{2} = b^{4}

改写方程式

- 16 u^{2} - 8 u + 3 = 0

解

- 16 u^{2} - 8 u + 3 = 0 : u = - \frac{3}{4}, u = \frac{1}{4}

- 16 u^{2} - 8 u + 3 = 0

使用求根公式求解

- 16 u^{2} - 8 u + 3 = 0

二次方程求根公式:

若

a = - 16, b = - 8, c = 3

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 16 ) \cdot 3}{2 ( - 16 )}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 16 ) \cdot 3}{2 ( - 16 )}

(- 8)^{2} - 4 (- 16) \cdot 3 = 16

(- 8)^{2} - 4 (- 16) \cdot 3

使用法则

- (- a) = a

= (- 8)^{2} + 4 \cdot 16 \cdot 3

使用指数法则:

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

若

n

是偶数

(- 8)^{2} = 8^{2}

= 8^{2} + 4 \cdot 16 \cdot 3

数字相乘:

4 \cdot 16 \cdot 3 = 192

= 8^{2} + 192

8^{2} = 64

= 64 + 192

数字相加:

64 + 192 = 256

= 256

因式分解数字:

256 = 1 6^{2}

= 1 6^{2}

使用根式运算法则:

n a^{n} = a

1 6^{2} = 16

= 16

u_{1, 2} = \frac{- ( - 8 ) \pm 16}{2 ( - 16 )}

将解分隔开

u_{1} = \frac{- ( - 8 ) + 16}{2 ( - 16 )}, u_{2} = \frac{- ( - 8 ) - 16}{2 ( - 16 )}

u = \frac{- ( - 8 ) + 16}{2 ( - 16 )} : - \frac{3}{4}

\frac{- ( - 8 ) + 16}{2 ( - 16 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{8 + 16}{- 2 \cdot 16}

数字相加:

8 + 16 = 24

= \frac{24}{- 2 \cdot 16}

数字相乘:

2 \cdot 16 = 32

= \frac{24}{- 32}

使用分式法则:

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

= - \frac{24}{32}

约分:

8

= - \frac{3}{4}

u = \frac{- ( - 8 ) - 16}{2 ( - 16 )} : \frac{1}{4}

\frac{- ( - 8 ) - 16}{2 ( - 16 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{8 - 16}{- 2 \cdot 16}

数字相减:

8 - 16 = - 8

= \frac{- 8}{- 2 \cdot 16}

数字相乘:

2 \cdot 16 = 32

= \frac{- 8}{- 32}

使用分式法则:

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

= \frac{8}{32}

约分:

8

= \frac{1}{4}

二次方程组的解是:

u = - \frac{3}{4}, u = \frac{1}{4}

u = - \frac{3}{4}, u = \frac{1}{4}

代回

u = b^{2}

，求解

b

解

b^{2} = - \frac{3}{4} : b \in R

无解

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

x^{2}

在

x

内不能为负

\in R

b \in R 无解

解

b^{2} = \frac{1}{4} : b = \frac{1}{2}, b = - \frac{1}{2}

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

对于

x^{2} = f (a)

解为

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

b = \frac{1}{4}, b = - \frac{1}{4}

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

\frac{1}{4}

使用根式运算法则:

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

= \frac{1}{4}

使用根式运算法则:

1 = 1

1 = 1

= \frac{1}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

a^{2} = a, a \geq 0

2^{2} = 2

= 2

= \frac{1}{2}

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

- \frac{1}{4}

使用根式运算法则:

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

= - \frac{1}{4}

使用根式运算法则:

1 = 1

1 = 1

= - \frac{1}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

a^{2} = a, a \geq 0

2^{2} = 2

= 2

= - \frac{1}{2}

b = \frac{1}{2}, b = - \frac{1}{2}

解为

b = \frac{1}{2}, b = - \frac{1}{2}

b = \frac{1}{2}, b = - \frac{1}{2}

验证解

找到无定义的点（奇点）:

b = 0

取

(- \frac{3}{4 b})^{2} - b^{2}

的分母，令其等于零

解

4 b = 0 : b = 0

4 b = 0

两边除以

4

4 b = 0

两边除以

4

\frac{4 b}{4} = \frac{0}{4}

化简

b = 0

b = 0

以下点无定义

b = 0

将不在定义域的点与解相综合:

b = \frac{1}{2}, b = - \frac{1}{2}

将解

b = \frac{1}{2}, b = - \frac{1}{2}

代入

2 ab = - \frac{3}{2}

对于

2 ab = - \frac{3}{2}

，用

\frac{1}{2}

替代

b : a = - \frac{3}{2}

对于

2 ab = - \frac{3}{2}

，用

\frac{1}{2}

替代

b

2 a \frac{1}{2} = - \frac{3}{2}

解

2 a \frac{1}{2} = - \frac{3}{2} : a = - \frac{3}{2}

2 a \frac{1}{2} = - \frac{3}{2}

分式相乘:

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

\frac{1 \cdot 2}{2} a = - \frac{3}{2}

约分:

2

a \cdot 1 = - \frac{3}{2}

乘以:

a \cdot 1 = a

a = - \frac{3}{2}

对于

2 ab = - \frac{3}{2}

，用

- \frac{1}{2}

替代

b : a = \frac{3}{2}

对于

2 ab = - \frac{3}{2}

，用

- \frac{1}{2}

替代

b

2 a (- \frac{1}{2}) = - \frac{3}{2}

解

2 a (- \frac{1}{2}) = - \frac{3}{2} : a = \frac{3}{2}

2 a (- \frac{1}{2}) = - \frac{3}{2}

两边除以

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

2 a (- \frac{1}{2}) = - \frac{3}{2}

两边除以

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

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} = \frac{- \frac{3}{2}}{2 ( - \frac{1}{2} )}

化简

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} = \frac{- \frac{3}{2}}{2 ( - \frac{1}{2} )}

化简

\frac{2 a ( - \frac{1}{2} )}{2 ( - \frac{1}{2} )} : a

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

去除括号:

(- a) = - a

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

使用分式法则:

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

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

乘

2 a \frac{1}{2} : a

2 a \frac{1}{2}

分式相乘:

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

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

约分:

2

= 1 \cdot a

乘以:

1 \cdot a = a

= a

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

乘

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

= \frac{a}{1}

使用法则

\frac{a}{1} = a

= a

化简

\frac{- \frac{3}{2}}{2 ( - \frac{1}{2} )} : \frac{3}{2}

\frac{- \frac{3}{2}}{2 ( - \frac{1}{2} )}

去除括号:

(- a) = - a

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

使用分式法则:

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

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

使用分式法则:

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

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

数字相乘:

2 \cdot 2 = 4

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

乘

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

4 \cdot \frac{1}{2}

分式相乘:

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

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

数字相乘:

1 \cdot 4 = 4

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

= \frac{3}{2}

a = \frac{3}{2}

a = \frac{3}{2}

a = \frac{3}{2}

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

a = \frac{3}{2}, b = - \frac{1}{2}

的解:

真

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

代入

a = \frac{3}{2}, b = - \frac{1}{2}

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

整理后得

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

真

检验

a = - \frac{3}{2}, b = \frac{1}{2}

的解:

真

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

代入

a = - \frac{3}{2}, b = \frac{1}{2}

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

整理后得

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

真

将它们代入

2 ab = - \frac{3}{2}

检验解是否符合

去除与方程不符的解。

检验

a = \frac{3}{2}, b = - \frac{1}{2}

的解:

真

2 ab = - \frac{3}{2}

代入

a = \frac{3}{2}, b = - \frac{1}{2}

2 \cdot \frac{3}{2} (- \frac{1}{2}) = - \frac{3}{2}

整理后得

- \frac{3}{2} = - \frac{3}{2}

真

检验

a = - \frac{3}{2}, b = \frac{1}{2}

的解:

真

2 ab = - \frac{3}{2}

代入

a = - \frac{3}{2}, b = \frac{1}{2}

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

整理后得

- \frac{3}{2} = - \frac{3}{2}

真

因而，

a^{2} - b^{2} = \frac{1}{2}, 2 ab = - \frac{3}{2}

最后的解是

(a = - \frac{3}{2}, a = \frac{3}{2}, b = \frac{1}{2} b = - \frac{1}{2})

u = a + bi

代回

u = - \frac{3}{2} + \frac{1}{2} i, u = \frac{3}{2} - \frac{1}{2} i

解为

u = \frac{3}{2} + \frac{1}{2} i, u = - \frac{3}{2} - \frac{1}{2} i, u = - \frac{3}{2} + \frac{1}{2} i, u = \frac{3}{2} - \frac{1}{2} i

u = sin (x)

代回

sin (x) = \frac{3}{2} + \frac{1}{2} i, sin (x) = - \frac{3}{2} - \frac{1}{2} i, sin (x) = - \frac{3}{2} + \frac{1}{2} i, sin (x) = \frac{3}{2} - \frac{1}{2} i

sin (x) = \frac{3}{2} + \frac{1}{2} i, sin (x) = - \frac{3}{2} - \frac{1}{2} i, sin (x) = - \frac{3}{2} + \frac{1}{2} i, sin (x) = \frac{3}{2} - \frac{1}{2} i

sin (x) = \frac{3}{2} + \frac{1}{2} i :

无解

sin (x) = \frac{3}{2} + \frac{1}{2} i

无解

sin (x) = - \frac{3}{2} - \frac{1}{2} i :

无解

sin (x) = - \frac{3}{2} - \frac{1}{2} i

无解

sin (x) = - \frac{3}{2} + \frac{1}{2} i :

无解

sin (x) = - \frac{3}{2} + \frac{1}{2} i

无解

sin (x) = \frac{3}{2} - \frac{1}{2} i :

无解

sin (x) = \frac{3}{2} - \frac{1}{2} i

无解

合并所有解

无解

合并所有解

x = 2 πn, x = π + 2 πn

因为方程对以下值无定义:

2 πn, π + 2 πn

x \in R 无解

作图

流行的例子

sin (x) = \frac{53}{54}

2sin^2(x)+sin(x)-3=0

2 sin^{2} (x) + sin (x) - 3 = 0

(2cos(x)-1)sin(x)=0

(2 cos (x) - 1) sin (x) = 0

solvefor u,x=4sin(u)

so l v e f or u, x = 4 sin (u)

2cos(θ)=sqrt(2),0<= θ<2pi

2 cos (θ) = 2, 0 \leq θ < 2 π