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受欢迎的三角函数 >

csch(x)= 5/12

解答

csch (x) = \frac{5}{12}

解答

x = ln (5)

+1

度数

x = 92.21399 \dots^{\circ}

求解步骤

csch (x) = \frac{5}{12}

使用三角恒等式改写

csch (x) = \frac{5}{12}

使用双曲函数恒等式:

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

\frac{2}{e ^{x} - e ^{- x}} = \frac{5}{12}

\frac{2}{e ^{x} - e ^{- x}} = \frac{5}{12}

\frac{2}{e ^{x} - e ^{- x}} = \frac{5}{12} : x = ln (5)

\frac{2}{e ^{x} - e ^{- x}} = \frac{5}{12}

使用分式交叉相乘: 若

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

则

a \cdot d = b \cdot c

2 \cdot 12 = (e^{x} - e^{- x}) \cdot 5

化简

24 = (e^{x} - e^{- x}) \cdot 5

使用指数运算法则

24 = (e^{x} - e^{- x}) \cdot 5

使用指数法则:

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

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

24 = (e^{x} - (e^{x})^{- 1}) \cdot 5

24 = (e^{x} - (e^{x})^{- 1}) \cdot 5

用

e^{x} = u

改写方程式

24 = (u - (u)^{- 1}) \cdot 5

解

24 = (u - u^{- 1}) \cdot 5 : u = 5, u = - \frac{1}{5}

24 = (u - u^{- 1}) \cdot 5

整理后得

24 = (u - \frac{1}{u}) \cdot 5

化简

(u - \frac{1}{u}) \cdot 5 : 5 (u - \frac{1}{u})

(u - \frac{1}{u}) \cdot 5

使用交换律:

(u - \frac{1}{u}) \cdot 5 = 5 (u - \frac{1}{u})

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

24 = 5 (u - \frac{1}{u})

展开

5 (u - \frac{1}{u}) : 5 u - \frac{5}{u}

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

使用分配律:

a (b - c) = ab - a c

a = 5, b = u, c = \frac{1}{u}

= 5 u - 5 \cdot \frac{1}{u}

5 \cdot \frac{1}{u} = \frac{5}{u}

5 \cdot \frac{1}{u}

分式相乘:

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

= \frac{1 \cdot 5}{u}

数字相乘:

1 \cdot 5 = 5

= \frac{5}{u}

= 5 u - \frac{5}{u}

24 = 5 u - \frac{5}{u}

在两边乘以

u

24 = 5 u - \frac{5}{u}

在两边乘以

u

24 u = 5 uu - \frac{5}{u} u

化简

24 u = 5 uu - \frac{5}{u} u

化简

5 uu : 5 u^{2}

5 uu

使用指数法则:

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

uu = u^{1 + 1}

= 5 u^{1 + 1}

数字相加:

1 + 1 = 2

= 5 u^{2}

化简

- \frac{5}{u} u : - 5

- \frac{5}{u} u

分式相乘:

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

= - \frac{5 u}{u}

约分:

u

= - 5

24 u = 5 u^{2} - 5

24 u = 5 u^{2} - 5

24 u = 5 u^{2} - 5

解

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

24 u = 5 u^{2} - 5

交换两边

5 u^{2} - 5 = 24 u

将

24 u

para o lado esquerdo

5 u^{2} - 5 = 24 u

两边减去

24 u

5 u^{2} - 5 - 24 u = 24 u - 24 u

化简

5 u^{2} - 5 - 24 u = 0

5 u^{2} - 5 - 24 u = 0

改写成标准形式

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

5 u^{2} - 24 u - 5 = 0

使用求根公式求解

5 u^{2} - 24 u - 5 = 0

二次方程求根公式:

若

a = 5, b = - 24, c = - 5

u_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 5 ( - 5 )}{2 \cdot 5}

u_{1, 2} = \frac{- ( - 24 ) \pm ( - 24 ) ^{2} - 4 \cdot 5 ( - 5 )}{2 \cdot 5}

(- 24)^{2} - 4 \cdot 5 (- 5) = 26

(- 24)^{2} - 4 \cdot 5 (- 5)

使用法则

- (- a) = a

= (- 24)^{2} + 4 \cdot 5 \cdot 5

使用指数法则:

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

若

n

是偶数

(- 24)^{2} = 2 4^{2}

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

数字相乘:

4 \cdot 5 \cdot 5 = 100

= 2 4^{2} + 100

2 4^{2} = 576

= 576 + 100

数字相加:

576 + 100 = 676

= 676

因式分解数字:

676 = 2 6^{2}

= 2 6^{2}

使用根式运算法则:

n a^{n} = a

2 6^{2} = 26

= 26

u_{1, 2} = \frac{- ( - 24 ) \pm 26}{2 \cdot 5}

将解分隔开

u_{1} = \frac{- ( - 24 ) + 26}{2 \cdot 5}, u_{2} = \frac{- ( - 24 ) - 26}{2 \cdot 5}

u = \frac{- ( - 24 ) + 26}{2 \cdot 5} : 5

\frac{- ( - 24 ) + 26}{2 \cdot 5}

使用法则

- (- a) = a

= \frac{24 + 26}{2 \cdot 5}

数字相加:

24 + 26 = 50

= \frac{50}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{50}{10}

数字相除:

\frac{50}{10} = 5

= 5

u = \frac{- ( - 24 ) - 26}{2 \cdot 5} : - \frac{1}{5}

\frac{- ( - 24 ) - 26}{2 \cdot 5}

使用法则

- (- a) = a

= \frac{24 - 26}{2 \cdot 5}

数字相减:

24 - 26 = - 2

= \frac{- 2}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{- 2}{10}

使用分式法则:

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

= - \frac{2}{10}

约分:

2

= - \frac{1}{5}

二次方程组的解是:

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

(u - u^{- 1}) 5

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}{5}

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

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}{5} : x \in R

无解

e^{x} = - \frac{1}{5}

a^{f (x)}

对于

x

不能为零或负值

\in R

x \in R 无解

x = ln (5)

验证解:

x = ln (5)

真

将它们代入

\frac{2}{e ^{x} - e ^{- x}} = \frac{5}{12}

检验解是否符合

去除与方程不符的解。

代入

x = ln (5) :

真

\frac{2}{e ^{l n (5)} - e ^{- l n (5)}} = \frac{5}{12}

\frac{2}{e ^{l n (5)} - e ^{- l n (5)}} = \frac{5}{12}

\frac{2}{e ^{l n (5)} - e ^{- l n (5)}}

e^{l n (5)} = 5

e^{l n (5)}

使用对数计算法则:

a^{l o g_{a} (b)} = b

= 5

e^{- l n (5)} = 5^{- 1}

e^{- l n (5)}

使用指数法则:

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

= (e^{l n (5)})^{- 1}

使用对数计算法则:

a^{l o g_{a} (b)} = b

e^{l n (5)} = 5

= 5^{- 1}

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

化简

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

使用指数法则:

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

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

化简

5 - \frac{1}{5} : \frac{24}{5}

5 - \frac{1}{5}

将项转换为分式:

5 = \frac{5 \cdot 5}{5}

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

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{5 \cdot 5 - 1}{5}

5 \cdot 5 - 1 = 24

5 \cdot 5 - 1

数字相乘:

5 \cdot 5 = 25

= 25 - 1

数字相减:

25 - 1 = 24

= 24

= \frac{24}{5}

= \frac{2}{\frac{24}{5}}

使用分式法则:

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

= \frac{2 \cdot 5}{24}

数字相乘:

2 \cdot 5 = 10

= \frac{10}{24}

约分:

2

= \frac{5}{12}

= \frac{5}{12}

\frac{5}{12} = \frac{5}{12}

真

解是

x = ln (5)

x = ln (5)

作图

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