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受欢迎的三角函数 >

sinh^2(x)+3tanh^2(x)=4

解答

sinh^{2} (x) + 3 tanh^{2} (x) = 4

解答

x = \frac{1}{2} ln (\frac{5.05103 E 15}{5.0 E 16}), x = \frac{1}{2} ln (\frac{4.94949 E 15}{5.0 E 14})

+1

度数

x = - 65.67332 \dots^{\circ}, x = 65.67332 \dots^{\circ}

求解步骤

sinh^{2} (x) + 3 tanh^{2} (x) = 4

使用三角恒等式改写

sinh^{2} (x) + 3 tanh^{2} (x) = 4

使用双曲函数恒等式:

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

(\frac{e ^{x} - e ^{- x}}{2})^{2} + 3 tanh^{2} (x) = 4

使用双曲函数恒等式:

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

(\frac{e ^{x} - e ^{- x}}{2})^{2} + 3 (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})^{2} = 4

(\frac{e ^{x} - e ^{- x}}{2})^{2} + 3 (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})^{2} = 4

(\frac{e ^{x} - e ^{- x}}{2})^{2} + 3 (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})^{2} = 4 : x = \frac{1}{2} ln (\frac{5.05103 E 15}{5.0 E 16}), x = \frac{1}{2} ln (\frac{4.94949 E 15}{5.0 E 14})

(\frac{e ^{x} - e ^{- x}}{2})^{2} + 3 (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})^{2} = 4

使用指数运算法则

(\frac{e ^{x} - e ^{- x}}{2})^{2} + 3 (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})^{2} = 4

使用指数法则:

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

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

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

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

用

e^{x} = u

改写方程式

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

解

(\frac{u - u ^{- 1}}{2})^{2} + 3 (\frac{u - u ^{- 1}}{u + u ^{- 1}})^{2} = 4 : u = \frac{5.05103 E 15}{5.0 E 16}, u = - \frac{5.05103 E 15}{5.0 E 16}, u = \frac{4.94949 E 15}{5.0 E 14}, u = - \frac{4.94949 E 15}{5.0 E 14}

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

整理后得

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

乘以最小公倍数

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

找到

4 u^{2}, (u^{2} + 1)^{2}

的最小公倍数:

4 u^{2} (u^{2} + 1)^{2}

4 u^{2}, (u^{2} + 1)^{2}

最小公倍数 (LCM)

计算出由出现在

4 u^{2}

或

(u^{2} + 1)^{2}

中的因子组成的表达式

= 4 u^{2} (u^{2} + 1)^{2}

乘以最小公倍数=

4 u^{2} (u^{2} + 1)^{2}

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

化简

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

化简

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

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

分式相乘:

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

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

约分:

4

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

约分:

u^{2}

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

化简

\frac{3 ( u ^{2} - 1 ) ^{2}}{( u ^{2} + 1 ) ^{2}} \cdot 4 u^{2} (u^{2} + 1)^{2} : 12 u^{2} (u^{2} - 1)^{2}

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

分式相乘:

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

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

约分:

(u^{2} + 1)^{2}

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

数字相乘:

3 \cdot 4 = 12

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

化简

4 \cdot 4 u^{2} (u^{2} + 1)^{2} : 16 u^{2} (u^{2} + 1)^{2}

4 \cdot 4 u^{2} (u^{2} + 1)^{2}

数字相乘:

4 \cdot 4 = 16

= 16 u^{2} (u^{2} + 1)^{2}

(u^{2} - 1)^{2} (u^{2} + 1)^{2} + 12 u^{2} (u^{2} - 1)^{2} = 16 u^{2} (u^{2} + 1)^{2}

(u^{2} - 1)^{2} (u^{2} + 1)^{2} + 12 u^{2} (u^{2} - 1)^{2} = 16 u^{2} (u^{2} + 1)^{2}

(u^{2} - 1)^{2} (u^{2} + 1)^{2} + 12 u^{2} (u^{2} - 1)^{2} = 16 u^{2} (u^{2} + 1)^{2}

解

(u^{2} - 1)^{2} (u^{2} + 1)^{2} + 12 u^{2} (u^{2} - 1)^{2} = 16 u^{2} (u^{2} + 1)^{2} : u = \frac{5.05103 E 15}{5.0 E 16}, u = - \frac{5.05103 E 15}{5.0 E 16}, u = \frac{4.94949 E 15}{5.0 E 14}, u = - \frac{4.94949 E 15}{5.0 E 14}

(u^{2} - 1)^{2} (u^{2} + 1)^{2} + 12 u^{2} (u^{2} - 1)^{2} = 16 u^{2} (u^{2} + 1)^{2}

展开

(u^{2} - 1)^{2} (u^{2} + 1)^{2} + 12 u^{2} (u^{2} - 1)^{2} : u^{8} + 12 u^{6} - 26 u^{4} + 12 u^{2} + 1

(u^{2} - 1)^{2} (u^{2} + 1)^{2} + 12 u^{2} (u^{2} - 1)^{2}

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

(u^{2} - 1)^{2} (u^{2} + 1)^{2}

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

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

使用完全平方公式:

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

a = u^{2}, b = 1

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

化简

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

(u^{2})^{2} - 2 u^{2} \cdot 1 + 1^{2}

使用法则

1^{a} = 1

1^{2} = 1

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

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

(u^{2})^{2}

使用指数法则:

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

= u^{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= u^{4}

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

2 u^{2} \cdot 1

数字相乘:

2 \cdot 1 = 2

= 2 u^{2}

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

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

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

(u^{2} + 1)^{2} = u^{4} + 2 u^{2} + 1

(u^{2} + 1)^{2}

使用完全平方公式:

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

a = u^{2}, b = 1

= (u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2}

化简

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

(u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2}

使用法则

1^{a} = 1

1^{2} = 1

= (u^{2})^{2} + 2 \cdot 1 \cdot u^{2} + 1

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

(u^{2})^{2}

使用指数法则:

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

= u^{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= u^{4}

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

2 u^{2} \cdot 1

数字相乘:

2 \cdot 1 = 2

= 2 u^{2}

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

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

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

12 u^{2} (u^{2} - 1)^{2} = 12 u^{2} (u^{4} - 2 u^{2} + 1)

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

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

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

使用完全平方公式:

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

a = u^{2}, b = 1

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

化简

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

(u^{2})^{2} - 2 u^{2} \cdot 1 + 1^{2}

使用法则

1^{a} = 1

1^{2} = 1

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

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

(u^{2})^{2}

使用指数法则:

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

= u^{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= u^{4}

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

2 u^{2} \cdot 1

数字相乘:

2 \cdot 1 = 2

= 2 u^{2}

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

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

= 12 u^{2} (u^{4} - 2 u^{2} + 1)

= (u^{4} - 2 u^{2} + 1) (u^{4} + 2 u^{2} + 1) + 12 u^{2} (u^{4} - 2 u^{2} + 1)

乘开

(u^{4} - 2 u^{2} + 1) (u^{4} + 2 u^{2} + 1) : u^{8} - 2 u^{4} + 1

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

打开括号

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

使用加减运算法则

+ (- a) = - a

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

化简

u^{4} u^{4} + 2 u^{4} u^{2} + 1 \cdot u^{4} - 2 u^{4} u^{2} - 2 \cdot 2 u^{2} u^{2} - 2 \cdot 1 \cdot u^{2} + 1 \cdot u^{4} + 1 \cdot 2 u^{2} + 1 \cdot 1 : u^{8} - 2 u^{4} + 1

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

对同类项分组

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

同类项相加:

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

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

同类项相加:

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

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

同类项相加:

- 2 \cdot 1 \cdot u^{2} + 1 \cdot 2 u^{2} = 0

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

u^{4} u^{4} = u^{8}

u^{4} u^{4}

使用指数法则:

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

u^{4} u^{4} = u^{4 + 4}

= u^{4 + 4}

数字相加:

4 + 4 = 8

= u^{8}

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

2 \cdot 2 u^{2} u^{2}

数字相乘:

2 \cdot 2 = 4

= 4 u^{2} u^{2}

使用指数法则:

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

u^{2} u^{2} = u^{2 + 2}

= 4 u^{2 + 2}

数字相加:

2 + 2 = 4

= 4 u^{4}

1 \cdot 1 = 1

1 \cdot 1

数字相乘:

1 \cdot 1 = 1

= 1

= u^{8} + 2 u^{4} - 4 u^{4} + 1

同类项相加:

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

= u^{8} - 2 u^{4} + 1

= u^{8} - 2 u^{4} + 1

= u^{8} - 2 u^{4} + 1 + 12 u^{2} (u^{4} - 2 u^{2} + 1)

乘开

12 u^{2} (u^{4} - 2 u^{2} + 1) : 12 u^{6} - 24 u^{4} + 12 u^{2}

12 u^{2} (u^{4} - 2 u^{2} + 1)

打开括号

= 12 u^{2} u^{4} + 12 u^{2} (- 2 u^{2}) + 12 u^{2} \cdot 1

使用加减运算法则

+ (- a) = - a

= 12 u^{4} u^{2} - 12 \cdot 2 u^{2} u^{2} + 12 \cdot 1 \cdot u^{2}

化简

12 u^{4} u^{2} - 12 \cdot 2 u^{2} u^{2} + 12 \cdot 1 \cdot u^{2} : 12 u^{6} - 24 u^{4} + 12 u^{2}

12 u^{4} u^{2} - 12 \cdot 2 u^{2} u^{2} + 12 \cdot 1 \cdot u^{2}

12 u^{4} u^{2} = 12 u^{6}

12 u^{4} u^{2}

使用指数法则:

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

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

= 12 u^{4 + 2}

数字相加:

4 + 2 = 6

= 12 u^{6}

12 \cdot 2 u^{2} u^{2} = 24 u^{4}

12 \cdot 2 u^{2} u^{2}

数字相乘:

12 \cdot 2 = 24

= 24 u^{2} u^{2}

使用指数法则:

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

u^{2} u^{2} = u^{2 + 2}

= 24 u^{2 + 2}

数字相加:

2 + 2 = 4

= 24 u^{4}

12 \cdot 1 \cdot u^{2} = 12 u^{2}

12 \cdot 1 \cdot u^{2}

数字相乘:

12 \cdot 1 = 12

= 12 u^{2}

= 12 u^{6} - 24 u^{4} + 12 u^{2}

= 12 u^{6} - 24 u^{4} + 12 u^{2}

= u^{8} - 2 u^{4} + 1 + 12 u^{6} - 24 u^{4} + 12 u^{2}

化简

u^{8} - 2 u^{4} + 1 + 12 u^{6} - 24 u^{4} + 12 u^{2} : u^{8} + 12 u^{6} - 26 u^{4} + 12 u^{2} + 1

u^{8} - 2 u^{4} + 1 + 12 u^{6} - 24 u^{4} + 12 u^{2}

对同类项分组

= u^{8} + 12 u^{6} - 2 u^{4} - 24 u^{4} + 12 u^{2} + 1

同类项相加:

- 2 u^{4} - 24 u^{4} = - 26 u^{4}

= u^{8} + 12 u^{6} - 26 u^{4} + 12 u^{2} + 1

= u^{8} + 12 u^{6} - 26 u^{4} + 12 u^{2} + 1

展开

16 u^{2} (u^{2} + 1)^{2} : 16 u^{6} + 32 u^{4} + 16 u^{2}

16 u^{2} (u^{2} + 1)^{2}

(u^{2} + 1)^{2} = u^{4} + 2 u^{2} + 1

(u^{2} + 1)^{2}

使用完全平方公式:

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

a = u^{2}, b = 1

= (u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2}

化简

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

(u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2}

使用法则

1^{a} = 1

1^{2} = 1

= (u^{2})^{2} + 2 \cdot 1 \cdot u^{2} + 1

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

(u^{2})^{2}

使用指数法则:

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

= u^{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= u^{4}

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

2 u^{2} \cdot 1

数字相乘:

2 \cdot 1 = 2

= 2 u^{2}

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

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

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

打开括号

= 16 u^{2} u^{4} + 16 u^{2} \cdot 2 u^{2} + 16 u^{2} \cdot 1

= 16 u^{4} u^{2} + 16 \cdot 2 u^{2} u^{2} + 16 \cdot 1 \cdot u^{2}

化简

16 u^{4} u^{2} + 16 \cdot 2 u^{2} u^{2} + 16 \cdot 1 \cdot u^{2} : 16 u^{6} + 32 u^{4} + 16 u^{2}

16 u^{4} u^{2} + 16 \cdot 2 u^{2} u^{2} + 16 \cdot 1 \cdot u^{2}

16 u^{4} u^{2} = 16 u^{6}

16 u^{4} u^{2}

使用指数法则:

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

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

= 16 u^{4 + 2}

数字相加:

4 + 2 = 6

= 16 u^{6}

16 \cdot 2 u^{2} u^{2} = 32 u^{4}

16 \cdot 2 u^{2} u^{2}

数字相乘:

16 \cdot 2 = 32

= 32 u^{2} u^{2}

使用指数法则:

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

u^{2} u^{2} = u^{2 + 2}

= 32 u^{2 + 2}

数字相加:

2 + 2 = 4

= 32 u^{4}

16 \cdot 1 \cdot u^{2} = 16 u^{2}

16 \cdot 1 \cdot u^{2}

数字相乘:

16 \cdot 1 = 16

= 16 u^{2}

= 16 u^{6} + 32 u^{4} + 16 u^{2}

= 16 u^{6} + 32 u^{4} + 16 u^{2}

u^{8} + 12 u^{6} - 26 u^{4} + 12 u^{2} + 1 = 16 u^{6} + 32 u^{4} + 16 u^{2}

将

16 u^{2}

para o lado esquerdo

u^{8} + 12 u^{6} - 26 u^{4} + 12 u^{2} + 1 = 16 u^{6} + 32 u^{4} + 16 u^{2}

两边减去

16 u^{2}

u^{8} + 12 u^{6} - 26 u^{4} + 12 u^{2} + 1 - 16 u^{2} = 16 u^{6} + 32 u^{4} + 16 u^{2} - 16 u^{2}

化简

u^{8} + 12 u^{6} - 26 u^{4} - 4 u^{2} + 1 = 16 u^{6} + 32 u^{4}

u^{8} + 12 u^{6} - 26 u^{4} - 4 u^{2} + 1 = 16 u^{6} + 32 u^{4}

将

32 u^{4}

para o lado esquerdo

u^{8} + 12 u^{6} - 26 u^{4} - 4 u^{2} + 1 = 16 u^{6} + 32 u^{4}

两边减去

32 u^{4}

u^{8} + 12 u^{6} - 26 u^{4} - 4 u^{2} + 1 - 32 u^{4} = 16 u^{6} + 32 u^{4} - 32 u^{4}

化简

u^{8} + 12 u^{6} - 58 u^{4} - 4 u^{2} + 1 = 16 u^{6}

u^{8} + 12 u^{6} - 58 u^{4} - 4 u^{2} + 1 = 16 u^{6}

将

16 u^{6}

para o lado esquerdo

u^{8} + 12 u^{6} - 58 u^{4} - 4 u^{2} + 1 = 16 u^{6}

两边减去

16 u^{6}

u^{8} + 12 u^{6} - 58 u^{4} - 4 u^{2} + 1 - 16 u^{6} = 16 u^{6} - 16 u^{6}

化简

u^{8} - 4 u^{6} - 58 u^{4} - 4 u^{2} + 1 = 0

u^{8} - 4 u^{6} - 58 u^{4} - 4 u^{2} + 1 = 0

用

v = u^{2}, v^{2} = u^{4}, v^{3} = u^{6}

和

v^{4} = u^{8}

改写方程式

v^{4} - 4 v^{3} - 58 v^{2} - 4 v + 1 = 0

解

v^{4} - 4 v^{3} - 58 v^{2} - 4 v + 1 = 0 : v \approx 0.10102 \dots, v \approx - 0.17157 \dots, v \approx - 5.82842 \dots, v \approx 9.89897 \dots

v^{4} - 4 v^{3} - 58 v^{2} - 4 v + 1 = 0

使用牛顿-拉弗森方法找到

v^{4} - 4 v^{3} - 58 v^{2} - 4 v + 1 = 0

的一个解:

v \approx 0.10102 \dots

v^{4} - 4 v^{3} - 58 v^{2} - 4 v + 1 = 0

牛顿-拉弗森近似法定义

f (v) = v^{4} - 4 v^{3} - 58 v^{2} - 4 v + 1

找到

f^{^{'}} (v) : 4 v^{3} - 12 v^{2} - 116 v - 4

\frac{d}{d v} (v^{4} - 4 v^{3} - 58 v^{2} - 4 v + 1)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d v} (v^{4}) - \frac{d}{d v} (4 v^{3}) - \frac{d}{d v} (58 v^{2}) - \frac{d}{d v} (4 v) + \frac{d}{d v} (1)

\frac{d}{d v} (v^{4}) = 4 v^{3}

\frac{d}{d v} (v^{4})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 4 v^{4 - 1}

化简

= 4 v^{3}

\frac{d}{d v} (4 v^{3}) = 12 v^{2}

\frac{d}{d v} (4 v^{3})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 4 \frac{d}{d v} (v^{3})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 4 \cdot 3 v^{3 - 1}

化简

= 12 v^{2}

\frac{d}{d v} (58 v^{2}) = 116 v

\frac{d}{d v} (58 v^{2})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 58 \frac{d}{d v} (v^{2})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 58 \cdot 2 v^{2 - 1}

化简

= 116 v

\frac{d}{d v} (4 v) = 4

\frac{d}{d v} (4 v)

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 4 \frac{d v}{d v}

使用常见微分定则:

\frac{d v}{d v} = 1

= 4 \cdot 1

化简

= 4

\frac{d}{d v} (1) = 0

\frac{d}{d v} (1)

常数微分:

\frac{d}{d x} (a) = 0

= 0

= 4 v^{3} - 12 v^{2} - 116 v - 4 + 0

化简

= 4 v^{3} - 12 v^{2} - 116 v - 4

令

v_{0} = 0

计算

v_{n + 1}

至

Δ v_{n + 1} < 0.000001

v_{1} = 0.25 : Δ v_{1} = 0.25

f (v_{0}) = 0^{4} - 4 \cdot 0^{3} - 58 \cdot 0^{2} - 4 \cdot 0 + 1 = 1

f^{^{'}} (v_{0}) = 4 \cdot 0^{3} - 12 \cdot 0^{2} - 116 \cdot 0 - 4 = - 4

v_{1} = 0.25

Δ v_{1} = ∣ 0.25 - 0 ∣ = 0.25

Δ v_{1} = 0.25

v_{2} = 0.14065 \dots : Δ v_{2} = 0.10934 \dots

f (v_{1}) = 0.2 5^{4} - 4 \cdot 0.2 5^{3} - 58 \cdot 0.2 5^{2} - 4 \cdot 0.25 + 1 = - 3.68359375

f^{^{'}} (v_{1}) = 4 \cdot 0.2 5^{3} - 12 \cdot 0.2 5^{2} - 116 \cdot 0.25 - 4 = - 33.6875

v_{2} = 0.14065 \dots

Δ v_{2} = ∣ 0.14065 \dots - 0.25 ∣ = 0.10934 \dots

Δ v_{2} = 0.10934 \dots

v_{3} = 0.10556 \dots : Δ v_{3} = 0.03508 \dots

f (v_{2}) = 0.14065 \dots^{4} - 4 \cdot 0.14065 \dots^{3} - 58 \cdot 0.14065 \dots^{2} - 4 \cdot 0.14065 \dots + 1 = - 0.72080 \dots

f^{^{'}} (v_{2}) = 4 \cdot 0.14065 \dots^{3} - 12 \cdot 0.14065 \dots^{2} - 116 \cdot 0.14065 \dots - 4 = - 20.54213 \dots

v_{3} = 0.10556 \dots

Δ v_{3} = ∣ 0.10556 \dots - 0.14065 \dots ∣ = 0.03508 \dots

Δ v_{3} = 0.03508 \dots

v_{4} = 0.10109 \dots : Δ v_{4} = 0.00446 \dots

f (v_{3}) = 0.10556 \dots^{4} - 4 \cdot 0.10556 \dots^{3} - 58 \cdot 0.10556 \dots^{2} - 4 \cdot 0.10556 \dots + 1 = - 0.07319 \dots

f^{^{'}} (v_{3}) = 4 \cdot 0.10556 \dots^{3} - 12 \cdot 0.10556 \dots^{2} - 116 \cdot 0.10556 \dots - 4 = - 16.37457 \dots

v_{4} = 0.10109 \dots

Δ v_{4} = ∣ 0.10109 \dots - 0.10556 \dots ∣ = 0.00446 \dots

Δ v_{4} = 0.00446 \dots

v_{5} = 0.10102 \dots : Δ v_{5} = 0.00007 \dots

f (v_{4}) = 0.10109 \dots^{4} - 4 \cdot 0.10109 \dots^{3} - 58 \cdot 0.10109 \dots^{2} - 4 \cdot 0.10109 \dots + 1 = - 0.00118 \dots

f^{^{'}} (v_{4}) = 4 \cdot 0.10109 \dots^{3} - 12 \cdot 0.10109 \dots^{2} - 116 \cdot 0.10109 \dots - 4 = - 15.84554 \dots

v_{5} = 0.10102 \dots

Δ v_{5} = ∣ 0.10102 \dots - 0.10109 \dots ∣ = 0.00007 \dots

Δ v_{5} = 0.00007 \dots

v_{6} = 0.10102 \dots : Δ v_{6} = 2.08017 E - 8

f (v_{5}) = 0.10102 \dots^{4} - 4 \cdot 0.10102 \dots^{3} - 58 \cdot 0.10102 \dots^{2} - 4 \cdot 0.10102 \dots + 1 = - 3.29431 E - 7

f^{^{'}} (v_{5}) = 4 \cdot 0.10102 \dots^{3} - 12 \cdot 0.10102 \dots^{2} - 116 \cdot 0.10102 \dots - 4 = - 15.83672 \dots

v_{6} = 0.10102 \dots

Δ v_{6} = ∣ 0.10102 \dots - 0.10102 \dots ∣ = 2.08017 E - 8

Δ v_{6} = 2.08017 E - 8

v \approx 0.10102 \dots

使用长除法

Eq u a t i o n 0 : \frac{v ^{4} - 4 v ^{3} - 58 v ^{2} - 4 v + 1}{v - 0.10102 \dots} = v^{3} - 3.89897 \dots v^{2} - 58.39387 \dots v - 9.89897 \dots

v^{3} - 3.89897 \dots v^{2} - 58.39387 \dots v - 9.89897 \dots \approx 0

使用牛顿-拉弗森方法找到

v^{3} - 3.89897 \dots v^{2} - 58.39387 \dots v - 9.89897 \dots = 0

的一个解:

v \approx - 0.17157 \dots

v^{3} - 3.89897 \dots v^{2} - 58.39387 \dots v - 9.89897 \dots = 0

牛顿-拉弗森近似法定义

f (v) = v^{3} - 3.89897 \dots v^{2} - 58.39387 \dots v - 9.89897 \dots

找到

f^{^{'}} (v) : 3 v^{2} - 7.79795 \dots v - 58.39387 \dots

\frac{d}{d v} (v^{3} - 3.89897 \dots v^{2} - 58.39387 \dots v - 9.89897 \dots)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d v} (v^{3}) - \frac{d}{d v} (3.89897 \dots v^{2}) - \frac{d}{d v} (58.39387 \dots v) - \frac{d}{d v} (9.89897 \dots)

\frac{d}{d v} (v^{3}) = 3 v^{2}

\frac{d}{d v} (v^{3})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 3 v^{3 - 1}

化简

= 3 v^{2}

\frac{d}{d v} (3.89897 \dots v^{2}) = 7.79795 \dots v

\frac{d}{d v} (3.89897 \dots v^{2})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 3.89897 \dots \frac{d}{d v} (v^{2})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 3.89897 \dots \cdot 2 v^{2 - 1}

化简

= 7.79795 \dots v

\frac{d}{d v} (58.39387 \dots v) = 58.39387 \dots

\frac{d}{d v} (58.39387 \dots v)

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 58.39387 \dots \frac{d v}{d v}

使用常见微分定则:

\frac{d v}{d v} = 1

= 58.39387 \dots \cdot 1

化简

= 58.39387 \dots

\frac{d}{d v} (9.89897 \dots) = 0

\frac{d}{d v} (9.89897 \dots)

常数微分:

\frac{d}{d x} (a) = 0

= 0

= 3 v^{2} - 7.79795 \dots v - 58.39387 \dots - 0

化简

= 3 v^{2} - 7.79795 \dots v - 58.39387 \dots

令

v_{0} = 0

计算

v_{n + 1}

至

Δ v_{n + 1} < 0.000001

v_{1} = - 0.16952 \dots : Δ v_{1} = 0.16952 \dots

f (v_{0}) = 0^{3} - 3.89897 \dots \cdot 0^{2} - 58.39387 \dots \cdot 0 - 9.89897 \dots = - 9.89897 \dots

f^{^{'}} (v_{0}) = 3 \cdot 0^{2} - 7.79795 \dots \cdot 0 - 58.39387 \dots = - 58.39387 \dots

v_{1} = - 0.16952 \dots

Δ v_{1} = ∣ - 0.16952 \dots - 0 ∣ = 0.16952 \dots

Δ v_{1} = 0.16952 \dots

v_{2} = - 0.17157 \dots : Δ v_{2} = 0.00205 \dots

f (v_{1}) = (- 0.16952 \dots)^{3} - 3.89897 \dots (- 0.16952 \dots)^{2} - 58.39387 \dots (- 0.16952 \dots) - 9.89897 \dots = - 0.11691 \dots

f^{^{'}} (v_{1}) = 3 (- 0.16952 \dots)^{2} - 7.79795 \dots (- 0.16952 \dots) - 58.39387 \dots = - 56.98574 \dots

v_{2} = - 0.17157 \dots

Δ v_{2} = ∣ - 0.17157 \dots - (- 0.16952 \dots) ∣ = 0.00205 \dots

Δ v_{2} = 0.00205 \dots

v_{3} = - 0.17157 \dots : Δ v_{3} = 3.25836 E - 7

f (v_{2}) = (- 0.17157 \dots)^{3} - 3.89897 \dots (- 0.17157 \dots)^{2} - 58.39387 \dots (- 0.17157 \dots) - 9.89897 \dots = - 0.00001 \dots

f^{^{'}} (v_{2}) = 3 (- 0.17157 \dots)^{2} - 7.79795 \dots (- 0.17157 \dots) - 58.39387 \dots = - 56.96764 \dots

v_{3} = - 0.17157 \dots

Δ v_{3} = ∣ - 0.17157 \dots - (- 0.17157 \dots) ∣ = 3.25836 E - 7

Δ v_{3} = 3.25836 E - 7

v \approx - 0.17157 \dots

使用长除法

Eq u a t i o n 0 : \frac{v ^{3} - 3.89897 \dots v ^{2} - 58.39387 \dots v - 9.89897 \dots}{v + 0.17157 \dots} = v^{2} - 4.07055 \dots v - 57.69548 \dots

v^{2} - 4.07055 \dots v - 57.69548 \dots \approx 0

使用牛顿-拉弗森方法找到

v^{2} - 4.07055 \dots v - 57.69548 \dots = 0

的一个解:

v \approx - 5.82842 \dots

v^{2} - 4.07055 \dots v - 57.69548 \dots = 0

牛顿-拉弗森近似法定义

f (v) = v^{2} - 4.07055 \dots v - 57.69548 \dots

找到

f^{^{'}} (v) : 2 v - 4.07055 \dots

\frac{d}{d v} (v^{2} - 4.07055 \dots v - 57.69548 \dots)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d v} (v^{2}) - \frac{d}{d v} (4.07055 \dots v) - \frac{d}{d v} (57.69548 \dots)

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

\frac{d}{d v} (v^{2})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 v^{2 - 1}

化简

= 2 v

\frac{d}{d v} (4.07055 \dots v) = 4.07055 \dots

\frac{d}{d v} (4.07055 \dots v)

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 4.07055 \dots \frac{d v}{d v}

使用常见微分定则:

\frac{d v}{d v} = 1

= 4.07055 \dots \cdot 1

化简

= 4.07055 \dots

\frac{d}{d v} (57.69548 \dots) = 0

\frac{d}{d v} (57.69548 \dots)

常数微分:

\frac{d}{d x} (a) = 0

= 0

= 2 v - 4.07055 \dots - 0

化简

= 2 v - 4.07055 \dots

令

v_{0} = - 5

计算

v_{n + 1}

至

Δ v_{n + 1} < 0.000001

v_{1} = - 5.87720 \dots : Δ v_{1} = 0.87720 \dots

f (v_{0}) = (- 5)^{2} - 4.07055 \dots (- 5) - 57.69548 \dots = - 12.34271 \dots

f^{^{'}} (v_{0}) = 2 (- 5) - 4.07055 \dots = - 14.07055 \dots

v_{1} = - 5.87720 \dots

Δ v_{1} = ∣ - 5.87720 \dots - (- 5) ∣ = 0.87720 \dots

Δ v_{1} = 0.87720 \dots

v_{2} = - 5.82857 \dots : Δ v_{2} = 0.04862 \dots

f (v_{1}) = (- 5.87720 \dots)^{2} - 4.07055 \dots (- 5.87720 \dots) - 57.69548 \dots = 0.76948 \dots

f^{^{'}} (v_{1}) = 2 (- 5.87720 \dots) - 4.07055 \dots = - 15.82495 \dots

v_{2} = - 5.82857 \dots

Δ v_{2} = ∣ - 5.82857 \dots - (- 5.87720 \dots) ∣ = 0.04862 \dots

Δ v_{2} = 0.04862 \dots

v_{3} = - 5.82842 \dots : Δ v_{3} = 0.00015 \dots

f (v_{2}) = (- 5.82857 \dots)^{2} - 4.07055 \dots (- 5.82857 \dots) - 57.69548 \dots = 0.00236 \dots

f^{^{'}} (v_{2}) = 2 (- 5.82857 \dots) - 4.07055 \dots = - 15.72770 \dots

v_{3} = - 5.82842 \dots

Δ v_{3} = ∣ - 5.82842 \dots - (- 5.82857 \dots) ∣ = 0.00015 \dots

Δ v_{3} = 0.00015 \dots

v_{4} = - 5.82842 \dots : Δ v_{4} = 1.43694 E - 9

f (v_{3}) = (- 5.82842 \dots)^{2} - 4.07055 \dots (- 5.82842 \dots) - 57.69548 \dots = 2.25994 E - 8

f^{^{'}} (v_{3}) = 2 (- 5.82842 \dots) - 4.07055 \dots = - 15.72740 \dots

v_{4} = - 5.82842 \dots

Δ v_{4} = ∣ - 5.82842 \dots - (- 5.82842 \dots) ∣ = 1.43694 E - 9

Δ v_{4} = 1.43694 E - 9

v \approx - 5.82842 \dots

使用长除法

Eq u a t i o n 0 : \frac{v ^{2} - 4.07055 \dots v - 57.69548 \dots}{v + 5.82842 \dots} = v - 9.89897 \dots

v - 9.89897 \dots \approx 0

v \approx 9.89897 \dots

解为

v \approx 0.10102 \dots, v \approx - 0.17157 \dots, v \approx - 5.82842 \dots, v \approx 9.89897 \dots

v \approx 0.10102 \dots, v \approx - 0.17157 \dots, v \approx - 5.82842 \dots, v \approx 9.89897 \dots

代回

v = u^{2}

，求解

u

解

u^{2} = 0.10102 \dots : u = \frac{5.05103 E 15}{5.0 E 16}, u = - \frac{5.05103 E 15}{5.0 E 16}

u^{2} = 0.10102 \dots

对于

x^{2} = f (a)

解为

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

u = 0.10102 \dots, u = - 0.10102 \dots

0.10102 \dots = \frac{5.05103 E 15}{5.0 E 16}

0.10102 \dots

小数点后有几位数字就乘以和除以几次 10。

小数点右侧有

17

位，因此乘以和除以

2147483647

= \frac{1.0 E 17 \cdot 0.10102 \dots}{1.0 E 17}

数字相乘:

1.0 E 17 \cdot 0.10102 \dots = 1.01021 E 16

= \frac{1.01021 E 16}{1.0 E 17}

消掉

\frac{1.01021 E 16}{1.0 E 17} : \frac{5.05103 E 15}{5.0 E 16}

\frac{1.01021 E 16}{1.0 E 17}

因式分解数字:

1.01021 E 16 = 2 \cdot 5.05103 E 15

= \frac{2 \cdot 5.05103 E 15}{1.0 E 17}

因式分解数字:

1.0 E 17 = 2 \cdot 5.0 E 16

= \frac{2 \cdot 5.05103 E 15}{2 \cdot 5.0 E 16}

约分:

2

= \frac{5.05103 E 15}{5.0 E 16}

= \frac{5.05103 E 15}{5.0 E 16}

- 0.10102 \dots = - \frac{5.05103 E 15}{5.0 E 16}

- 0.10102 \dots

小数点后有几位数字就乘以和除以几次 10。

小数点右侧有

17

位，因此乘以和除以

2147483647

= - \frac{1.0 E 17 \cdot 0.10102 \dots}{1.0 E 17}

数字相乘:

1.0 E 17 \cdot 0.10102 \dots = 1.01021 E 16

= - \frac{1.01021 E 16}{1.0 E 17}

消掉

\frac{1.01021 E 16}{1.0 E 17} : \frac{5.05103 E 15}{5.0 E 16}

\frac{1.01021 E 16}{1.0 E 17}

因式分解数字:

1.01021 E 16 = 2 \cdot 5.05103 E 15

= \frac{2 \cdot 5.05103 E 15}{1.0 E 17}

因式分解数字:

1.0 E 17 = 2 \cdot 5.0 E 16

= \frac{2 \cdot 5.05103 E 15}{2 \cdot 5.0 E 16}

约分:

2

= \frac{5.05103 E 15}{5.0 E 16}

= - \frac{5.05103 E 15}{5.0 E 16}

u = \frac{5.05103 E 15}{5.0 E 16}, u = - \frac{5.05103 E 15}{5.0 E 16}

解

u^{2} = - 0.17157 \dots : u \in R

无解

u^{2} = - 0.17157 \dots

x^{2}

在

x

内不能为负

\in R

u \in R 无解

解

u^{2} = - 5.82842 \dots : u \in R

无解

u^{2} = - 5.82842 \dots

x^{2}

在

x

内不能为负

\in R

u \in R 无解

解

u^{2} = 9.89897 \dots : u = \frac{4.94949 E 15}{5.0 E 14}, u = - \frac{4.94949 E 15}{5.0 E 14}

u^{2} = 9.89897 \dots

对于

x^{2} = f (a)

解为

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

u = 9.89897 \dots, u = - 9.89897 \dots

9.89897 \dots = \frac{4.94949 E 15}{5.0 E 14}

9.89897 \dots

小数点后有几位数字就乘以和除以几次 10。

小数点右侧有

15

位，因此乘以和除以

2147483647

= \frac{1.0 E 15 \cdot 9.89897 \dots}{1.0 E 15}

数字相乘:

1.0 E 15 \cdot 9.89897 \dots = 9.89898 E 15

= \frac{9.89898 E 15}{1.0 E 15}

消掉

\frac{9.89898 E 15}{1.0 E 15} : \frac{4.94949 E 15}{5.0 E 14}

\frac{9.89898 E 15}{1.0 E 15}

因式分解数字:

9.89898 E 15 = 2 \cdot 4.94949 E 15

= \frac{2 \cdot 4.94949 E 15}{1.0 E 15}

因式分解数字:

1.0 E 15 = 2 \cdot 5.0 E 14

= \frac{2 \cdot 4.94949 E 15}{2 \cdot 5.0 E 14}

约分:

2

= \frac{4.94949 E 15}{5.0 E 14}

= \frac{4.94949 E 15}{5.0 E 14}

- 9.89897 \dots = - \frac{4.94949 E 15}{5.0 E 14}

- 9.89897 \dots

小数点后有几位数字就乘以和除以几次 10。

小数点右侧有

15

位，因此乘以和除以

2147483647

= - \frac{1.0 E 15 \cdot 9.89897 \dots}{1.0 E 15}

数字相乘:

1.0 E 15 \cdot 9.89897 \dots = 9.89898 E 15

= - \frac{9.89898 E 15}{1.0 E 15}

消掉

\frac{9.89898 E 15}{1.0 E 15} : \frac{4.94949 E 15}{5.0 E 14}

\frac{9.89898 E 15}{1.0 E 15}

因式分解数字:

9.89898 E 15 = 2 \cdot 4.94949 E 15

= \frac{2 \cdot 4.94949 E 15}{1.0 E 15}

因式分解数字:

1.0 E 15 = 2 \cdot 5.0 E 14

= \frac{2 \cdot 4.94949 E 15}{2 \cdot 5.0 E 14}

约分:

2

= \frac{4.94949 E 15}{5.0 E 14}

= - \frac{4.94949 E 15}{5.0 E 14}

u = \frac{4.94949 E 15}{5.0 E 14}, u = - \frac{4.94949 E 15}{5.0 E 14}

解为

u = \frac{5.05103 E 15}{5.0 E 16}, u = - \frac{5.05103 E 15}{5.0 E 16}, u = \frac{4.94949 E 15}{5.0 E 14}, u = - \frac{4.94949 E 15}{5.0 E 14}

u = \frac{5.05103 E 15}{5.0 E 16}, u = - \frac{5.05103 E 15}{5.0 E 16}, u = \frac{4.94949 E 15}{5.0 E 14}, u = - \frac{4.94949 E 15}{5.0 E 14}

验证解

找到无定义的点（奇点）:

u = 0

取

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

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{5.05103 E 15}{5.0 E 16}, u = - \frac{5.05103 E 15}{5.0 E 16}, u = \frac{4.94949 E 15}{5.0 E 14}, u = - \frac{4.94949 E 15}{5.0 E 14}

u = \frac{5.05103 E 15}{5.0 E 16}, u = - \frac{5.05103 E 15}{5.0 E 16}, u = \frac{4.94949 E 15}{5.0 E 14}, u = - \frac{4.94949 E 15}{5.0 E 14}

代回

u = e^{x}

，求解

x

解

e^{x} = \frac{5.05103 E 15}{5.0 E 16} : x = \frac{1}{2} ln (\frac{5.05103 E 15}{5.0 E 16})

e^{x} = \frac{5.05103 E 15}{5.0 E 16}

使用指数运算法则

e^{x} = \frac{5.05103 E 15}{5.0 E 16}

使用指数法则:

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

\frac{5.05103 E 15}{5.0 E 16} = (\frac{5.05103 E 15}{5.0 E 16})^{\frac{1}{2}}

e^{x} = (\frac{5.05103 E 15}{5.0 E 16})^{\frac{1}{2}}

若

f (x) = g (x)

，则

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

ln (e^{x}) = ln ((\frac{5.05103 E 15}{5.0 E 16})^{\frac{1}{2}})

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln ((\frac{5.05103 E 15}{5.0 E 16})^{\frac{1}{2}})

使用对数计算法则:

ln (x^{a}) = a \cdot ln (x)

ln ((\frac{5.05103 E 15}{5.0 E 16})^{\frac{1}{2}}) = \frac{1}{2} ln (\frac{5.05103 E 15}{5.0 E 16})

x = \frac{1}{2} ln (\frac{5.05103 E 15}{5.0 E 16})

x = \frac{1}{2} ln (\frac{5.05103 E 15}{5.0 E 16})

解

e^{x} = - \frac{5.05103 E 15}{5.0 E 16} : x \in R

无解

e^{x} = - \frac{5.05103 E 15}{5.0 E 16}

a^{f (x)}

对于

x

不能为零或负值

\in R

x \in R 无解

解

e^{x} = \frac{4.94949 E 15}{5.0 E 14} : x = \frac{1}{2} ln (\frac{4.94949 E 15}{5.0 E 14})

e^{x} = \frac{4.94949 E 15}{5.0 E 14}

使用指数运算法则

e^{x} = \frac{4.94949 E 15}{5.0 E 14}

使用指数法则:

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

\frac{4.94949 E 15}{5.0 E 14} = (\frac{4.94949 E 15}{5.0 E 14})^{\frac{1}{2}}

e^{x} = (\frac{4.94949 E 15}{5.0 E 14})^{\frac{1}{2}}

若

f (x) = g (x)

，则

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

ln (e^{x}) = ln ((\frac{4.94949 E 15}{5.0 E 14})^{\frac{1}{2}})

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln ((\frac{4.94949 E 15}{5.0 E 14})^{\frac{1}{2}})

使用对数计算法则:

ln (x^{a}) = a \cdot ln (x)

ln ((\frac{4.94949 E 15}{5.0 E 14})^{\frac{1}{2}}) = \frac{1}{2} ln (\frac{4.94949 E 15}{5.0 E 14})

x = \frac{1}{2} ln (\frac{4.94949 E 15}{5.0 E 14})

x = \frac{1}{2} ln (\frac{4.94949 E 15}{5.0 E 14})

解

e^{x} = - \frac{4.94949 E 15}{5.0 E 14} : x \in R

无解

e^{x} = - \frac{4.94949 E 15}{5.0 E 14}

a^{f (x)}

对于

x

不能为零或负值

\in R

x \in R 无解

x = \frac{1}{2} ln (\frac{5.05103 E 15}{5.0 E 16}), x = \frac{1}{2} ln (\frac{4.94949 E 15}{5.0 E 14})

x = \frac{1}{2} ln (\frac{5.05103 E 15}{5.0 E 16}), x = \frac{1}{2} ln (\frac{4.94949 E 15}{5.0 E 14})

作图

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