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受欢迎的三角函数 >

4cosh(x)-sinh(x)=8

解答

4 cosh (x) - sinh (x) = 8

解答

x = ln (5), x = - ln (3)

+1

度数

x = 92.21399 \dots^{\circ}, x = - 62.94584 \dots^{\circ}

求解步骤

4 cosh (x) - sinh (x) = 8

使用三角恒等式改写

4 cosh (x) - sinh (x) = 8

使用双曲函数恒等式:

sinh (x) = \frac{e ^{x} - e ^{- x}}{2}

4 cosh (x) - \frac{e ^{x} - e ^{- x}}{2} = 8

使用双曲函数恒等式:

cosh (x) = \frac{e ^{x} + e ^{- x}}{2}

4 \cdot \frac{e ^{x} + e ^{- x}}{2} - \frac{e ^{x} - e ^{- x}}{2} = 8

4 \cdot \frac{e ^{x} + e ^{- x}}{2} - \frac{e ^{x} - e ^{- x}}{2} = 8

4 \cdot \frac{e ^{x} + e ^{- x}}{2} - \frac{e ^{x} - e ^{- x}}{2} = 8 : x = ln (5), x = - ln (3)

4 \cdot \frac{e ^{x} + e ^{- x}}{2} - \frac{e ^{x} - e ^{- x}}{2} = 8

在两边乘以

2

4 \cdot \frac{e ^{x} + e ^{- x}}{2} \cdot 2 - \frac{e ^{x} - e ^{- x}}{2} \cdot 2 = 8 \cdot 2

化简

4 (e^{x} + e^{- x}) - (e^{x} - e^{- x}) = 16

使用指数运算法则

4 (e^{x} + e^{- x}) - (e^{x} - e^{- x}) = 16

使用指数法则:

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

e^{- x} = (e^{x})^{- 1}

4 (e^{x} + (e^{x})^{- 1}) - (e^{x} - (e^{x})^{- 1}) = 16

4 (e^{x} + (e^{x})^{- 1}) - (e^{x} - (e^{x})^{- 1}) = 16

用

e^{x} = u

改写方程式

4 (u + (u)^{- 1}) - (u - (u)^{- 1}) = 16

解

4 (u + u^{- 1}) - (u - u^{- 1}) = 16 : u = 5, u = \frac{1}{3}

4 (u + u^{- 1}) - (u - u^{- 1}) = 16

整理后得

4 (u + \frac{1}{u}) - (u - \frac{1}{u}) = 16

化简

- (u - \frac{1}{u}) : - u + \frac{1}{u}

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

打开括号

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

使用加减运算法则

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

= - u + \frac{1}{u}

4 (u + \frac{1}{u}) - u + \frac{1}{u} = 16

在两边乘以

u

4 (u + \frac{1}{u}) - u + \frac{1}{u} = 16

在两边乘以

u

4 (u + \frac{1}{u}) u - uu + \frac{1}{u} u = 16 u

化简

4 (u + \frac{1}{u}) u - uu + \frac{1}{u} u = 16 u

化简

- uu : - u^{2}

- uu

使用指数法则:

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

uu = u^{1 + 1}

= - u^{1 + 1}

数字相加:

1 + 1 = 2

= - u^{2}

化简

\frac{1}{u} u : 1

\frac{1}{u} u

分式相乘:

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

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

约分:

u

= 1

4 (u + \frac{1}{u}) u - u^{2} + 1 = 16 u

4 (u + \frac{1}{u}) u - u^{2} + 1 = 16 u

4 (u + \frac{1}{u}) u - u^{2} + 1 = 16 u

展开

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

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

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

乘开

4 u (u + \frac{1}{u}) : 4 u^{2} + 4

4 u (u + \frac{1}{u})

使用分配律:

a (b + c) = ab + a c

a = 4 u, b = u, c = \frac{1}{u}

= 4 uu + 4 u \frac{1}{u}

= 4 uu + 4 \cdot \frac{1}{u} u

化简

4 uu + 4 \cdot \frac{1}{u} u : 4 u^{2} + 4

4 uu + 4 \cdot \frac{1}{u} u

4 uu = 4 u^{2}

4 uu

使用指数法则:

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

uu = u^{1 + 1}

= 4 u^{1 + 1}

数字相加:

1 + 1 = 2

= 4 u^{2}

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

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

分式相乘:

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

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

约分:

u

= 1 \cdot 4

数字相乘:

1 \cdot 4 = 4

= 4

= 4 u^{2} + 4

= 4 u^{2} + 4

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

化简

4 u^{2} + 4 - u^{2} + 1 : 3 u^{2} + 5

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

对同类项分组

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

同类项相加:

4 u^{2} - u^{2} = 3 u^{2}

= 3 u^{2} + 4 + 1

数字相加:

4 + 1 = 5

= 3 u^{2} + 5

= 3 u^{2} + 5

3 u^{2} + 5 = 16 u

解

3 u^{2} + 5 = 16 u : u = 5, u = \frac{1}{3}

3 u^{2} + 5 = 16 u

将

16 u

para o lado esquerdo

3 u^{2} + 5 = 16 u

两边减去

16 u

3 u^{2} + 5 - 16 u = 16 u - 16 u

化简

3 u^{2} + 5 - 16 u = 0

3 u^{2} + 5 - 16 u = 0

改写成标准形式

a x^{2} + b x + c = 0

3 u^{2} - 16 u + 5 = 0

使用求根公式求解

3 u^{2} - 16 u + 5 = 0

二次方程求根公式:

若

a = 3, b = - 16, c = 5

u_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 3 \cdot 5}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 3 \cdot 5}{2 \cdot 3}

(- 16)^{2} - 4 \cdot 3 \cdot 5 = 14

(- 16)^{2} - 4 \cdot 3 \cdot 5

使用指数法则:

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

若

n

是偶数

(- 16)^{2} = 1 6^{2}

= 1 6^{2} - 4 \cdot 3 \cdot 5

数字相乘:

4 \cdot 3 \cdot 5 = 60

= 1 6^{2} - 60

1 6^{2} = 256

= 256 - 60

数字相减:

256 - 60 = 196

= 196

因式分解数字:

196 = 1 4^{2}

= 1 4^{2}

使用根式运算法则:

n a^{n} = a

1 4^{2} = 14

= 14

u_{1, 2} = \frac{- ( - 16 ) \pm 14}{2 \cdot 3}

将解分隔开

u_{1} = \frac{- ( - 16 ) + 14}{2 \cdot 3}, u_{2} = \frac{- ( - 16 ) - 14}{2 \cdot 3}

u = \frac{- ( - 16 ) + 14}{2 \cdot 3} : 5

\frac{- ( - 16 ) + 14}{2 \cdot 3}

使用法则

- (- a) = a

= \frac{16 + 14}{2 \cdot 3}

数字相加:

16 + 14 = 30

= \frac{30}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{30}{6}

数字相除:

\frac{30}{6} = 5

= 5

u = \frac{- ( - 16 ) - 14}{2 \cdot 3} : \frac{1}{3}

\frac{- ( - 16 ) - 14}{2 \cdot 3}

使用法则

- (- a) = a

= \frac{16 - 14}{2 \cdot 3}

数字相减:

16 - 14 = 2

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

数字相乘:

2 \cdot 3 = 6

= \frac{2}{6}

约分:

2

= \frac{1}{3}

二次方程组的解是:

u = 5, u = \frac{1}{3}

u = 5, u = \frac{1}{3}

验证解

找到无定义的点（奇点）:

u = 0

取

4 (u + u^{- 1}) - (u - u^{- 1})

的分母，令其等于零

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = 5, u = \frac{1}{3}

u = 5, u = \frac{1}{3}

代回

u = e^{x}

，求解

x

解

e^{x} = 5 : x = ln (5)

e^{x} = 5

使用指数运算法则

e^{x} = 5

若

f (x) = g (x)

，则

ln (f (x)) = ln (g (x))

ln (e^{x}) = ln (5)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (5)

x = ln (5)

解

e^{x} = \frac{1}{3} : x = - ln (3)

e^{x} = \frac{1}{3}

使用指数运算法则

e^{x} = \frac{1}{3}

若

f (x) = g (x)

，则

ln (f (x)) = ln (g (x))

ln (e^{x}) = ln (\frac{1}{3})

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (\frac{1}{3})

化简

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

ln (\frac{1}{3})

使用对数计算法则:

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

= - ln (3)

x = - ln (3)

x = - ln (3)

x = ln (5), x = - ln (3)

x = ln (5), x = - ln (3)

作图

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