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受欢迎的三角函数 >

sinh(x)= 6/5

解答

sinh (x) = \frac{6}{5}

解答

x = ln (\frac{6 + 61}{5})

+1

度数

x = 58.21097 \dots^{\circ}

求解步骤

sinh (x) = \frac{6}{5}

使用三角恒等式改写

sinh (x) = \frac{6}{5}

使用双曲函数恒等式:

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

\frac{e ^{x} - e ^{- x}}{2} = \frac{6}{5}

\frac{e ^{x} - e ^{- x}}{2} = \frac{6}{5}

\frac{e ^{x} - e ^{- x}}{2} = \frac{6}{5} : x = ln (\frac{6 + 61}{5})

\frac{e ^{x} - e ^{- x}}{2} = \frac{6}{5}

使用分式交叉相乘: 若

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

则

a \cdot d = b \cdot c

(e^{x} - e^{- x}) \cdot 5 = 2 \cdot 6

化简

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

使用指数运算法则

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

使用指数法则:

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

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

(e^{x} - (e^{x})^{- 1}) \cdot 5 = 12

(e^{x} - (e^{x})^{- 1}) \cdot 5 = 12

用

e^{x} = u

改写方程式

(u - (u)^{- 1}) \cdot 5 = 12

解

(u - u^{- 1}) \cdot 5 = 12 : u = \frac{6 + 61}{5}, u = \frac{6 - 61}{5}

(u - u^{- 1}) \cdot 5 = 12

整理后得

(u - \frac{1}{u}) \cdot 5 = 12

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})

5 (u - \frac{1}{u}) = 12

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}

5 u - \frac{5}{u} = 12

在两边乘以

u

5 u - \frac{5}{u} = 12

在两边乘以

u

5 uu - \frac{5}{u} u = 12 u

化简

5 uu - \frac{5}{u} u = 12 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

5 u^{2} - 5 = 12 u

5 u^{2} - 5 = 12 u

5 u^{2} - 5 = 12 u

解

5 u^{2} - 5 = 12 u : u = \frac{6 + 61}{5}, u = \frac{6 - 61}{5}

5 u^{2} - 5 = 12 u

将

12 u

para o lado esquerdo

5 u^{2} - 5 = 12 u

两边减去

12 u

5 u^{2} - 5 - 12 u = 12 u - 12 u

化简

5 u^{2} - 5 - 12 u = 0

5 u^{2} - 5 - 12 u = 0

改写成标准形式

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

5 u^{2} - 12 u - 5 = 0

使用求根公式求解

5 u^{2} - 12 u - 5 = 0

二次方程求根公式:

若

a = 5, b = - 12, c = - 5

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 5 ( - 5 )}{2 \cdot 5}

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 5 ( - 5 )}{2 \cdot 5}

(- 12)^{2} - 4 \cdot 5 (- 5) = 261

(- 12)^{2} - 4 \cdot 5 (- 5)

使用法则

- (- a) = a

= (- 12)^{2} + 4 \cdot 5 \cdot 5

使用指数法则:

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

若

n

是偶数

(- 12)^{2} = 1 2^{2}

= 1 2^{2} + 4 \cdot 5 \cdot 5

数字相乘:

4 \cdot 5 \cdot 5 = 100

= 1 2^{2} + 100

1 2^{2} = 144

= 144 + 100

数字相加:

144 + 100 = 244

= 244

244

质因数分解:

2^{2} \cdot 61

244

244

除以

2244 = 122 \cdot 2

= 2 \cdot 122

122

除以

2122 = 61 \cdot 2

= 2 \cdot 2 \cdot 61

2, 61

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 61

= 2^{2} \cdot 61

= 2^{2} \cdot 61

使用根式运算法则:

n ab = n a n b

= 61 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 261

u_{1, 2} = \frac{- ( - 12 ) \pm 2 61}{2 \cdot 5}

将解分隔开

u_{1} = \frac{- ( - 12 ) + 2 61}{2 \cdot 5}, u_{2} = \frac{- ( - 12 ) - 2 61}{2 \cdot 5}

u = \frac{- ( - 12 ) + 2 61}{2 \cdot 5} : \frac{6 + 61}{5}

\frac{- ( - 12 ) + 2 61}{2 \cdot 5}

使用法则

- (- a) = a

= \frac{12 + 2 61}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{12 + 2 61}{10}

分解

12 + 261 : 2 (6 + 61)

12 + 261

改写为

= 2 \cdot 6 + 261

因式分解出通项

2

= 2 (6 + 61)

= \frac{2 ( 6 + 61 )}{10}

约分:

2

= \frac{6 + 61}{5}

u = \frac{- ( - 12 ) - 2 61}{2 \cdot 5} : \frac{6 - 61}{5}

\frac{- ( - 12 ) - 2 61}{2 \cdot 5}

使用法则

- (- a) = a

= \frac{12 - 2 61}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{12 - 2 61}{10}

分解

12 - 261 : 2 (6 - 61)

12 - 261

改写为

= 2 \cdot 6 - 261

因式分解出通项

2

= 2 (6 - 61)

= \frac{2 ( 6 - 61 )}{10}

约分:

2

= \frac{6 - 61}{5}

二次方程组的解是:

u = \frac{6 + 61}{5}, u = \frac{6 - 61}{5}

u = \frac{6 + 61}{5}, u = \frac{6 - 61}{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 = \frac{6 + 61}{5}, u = \frac{6 - 61}{5}

u = \frac{6 + 61}{5}, u = \frac{6 - 61}{5}

代回

u = e^{x}

，求解

x

解

e^{x} = \frac{6 + 61}{5} : x = ln (\frac{6 + 61}{5})

e^{x} = \frac{6 + 61}{5}

使用指数运算法则

e^{x} = \frac{6 + 61}{5}

若

f (x) = g (x)

，则

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

ln (e^{x}) = ln (\frac{6 + 61}{5})

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (\frac{6 + 61}{5})

x = ln (\frac{6 + 61}{5})

解

e^{x} = \frac{6 - 61}{5} : x \in R

无解

e^{x} = \frac{6 - 61}{5}

a^{f (x)}

对于

x

不能为零或负值

\in R

x \in R 无解

x = ln (\frac{6 + 61}{5})

x = ln (\frac{6 + 61}{5})

作图

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