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受欢迎的三角函数 >

75/125 =(cosh(0.2x))/(cosh(0.4x))

解答

\frac{75}{125} = \frac{cosh ( 0.2 x )}{cosh ( 0.4 x )}

解答

x = ln (0.03402 \dots), x = ln (29.38731 \dots)

+1

度数

x = - 193.69199 \dots^{\circ}, x = 193.69199 \dots^{\circ}

求解步骤

\frac{75}{125} = \frac{cosh ( 0.2 x )}{cosh ( 0.4 x )}

交换两边

\frac{cosh ( 0.2 x )}{cosh ( 0.4 x )} = \frac{75}{125}

使用三角恒等式改写

\frac{cosh ( 0.2 x )}{cosh ( 0.4 x )} = \frac{75}{125}

使用双曲函数恒等式:

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

\frac{\frac{e ^{0.2 x} + e ^{- 0.2 x}}{2}}{\frac{e ^{0.4 x} + e ^{- 0.4 x}}{2}} = \frac{75}{125}

\frac{\frac{e ^{0.2 x} + e ^{- 0.2 x}}{2}}{\frac{e ^{0.4 x} + e ^{- 0.4 x}}{2}} = \frac{75}{125}

\frac{\frac{e ^{0.2 x} + e ^{- 0.2 x}}{2}}{\frac{e ^{0.4 x} + e ^{- 0.4 x}}{2}} = \frac{75}{125} : x = ln (0.03402 \dots), x = ln (29.38731 \dots)

\frac{\frac{e ^{0.2 x} + e ^{- 0.2 x}}{2}}{\frac{e ^{0.4 x} + e ^{- 0.4 x}}{2}} = \frac{75}{125}

使用分式交叉相乘: 若

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

则

a \cdot d = b \cdot c

\frac{e ^{0.2 x} + e ^{- 0.2 x}}{2} \cdot 125 = \frac{e ^{0.4 x} + e ^{- 0.4 x}}{2} \cdot 75

在两边乘以

2

\frac{e ^{0.2 x} + e ^{- 0.2 x}}{2} \cdot 125 \cdot 2 = \frac{e ^{0.4 x} + e ^{- 0.4 x}}{2} \cdot 75 \cdot 2

化简

125 (e^{0.2 x} + e^{- 0.2 x}) = 75 (e^{0.4 x} + e^{- 0.4 x})

使用指数运算法则

125 (e^{0.2 x} + e^{- 0.2 x}) = 75 (e^{0.4 x} + e^{- 0.4 x})

使用指数法则:

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

e^{0.2 x} = (e^{x})^{0.2}, e^{- 0.2 x} = (e^{x})^{- 0.2}, e^{0.4 x} = (e^{x})^{0.4}, e^{- 0.4 x} = (e^{x})^{- 0.4}

125 ((e^{x})^{0.2} + (e^{x})^{- 0.2}) = 75 ((e^{x})^{0.4} + (e^{x})^{- 0.4})

125 ((e^{x})^{0.2} + (e^{x})^{- 0.2}) = 75 ((e^{x})^{0.4} + (e^{x})^{- 0.4})

用

e^{x} = u

改写方程式

125 ((u)^{0.2} + (u)^{- 0.2}) = 75 ((u)^{0.4} + (u)^{- 0.4})

解

125 (u^{0.2} + u^{- 0.2}) = 75 (u^{0.4} + u^{- 0.4}) : u = 0.03402 \dots, u = 29.38731 \dots

125 (u^{0.2} + u^{- 0.2}) = 75 (u^{0.4} + u^{- 0.4})

展开

125 (u^{0.2} + u^{- 0.2}) : 125 u^{0.2} + \frac{125}{u ^{0.2}}

125 (u^{0.2} + u^{- 0.2})

使用指数法则:

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

= 125 (u^{0.2} + \frac{1}{u ^{0.2}})

使用分配律:

a (b + c) = ab + a c

a = 125, b = u^{0.2}, c = \frac{1}{u ^{0.2}}

= 125 u^{0.2} + 125 \cdot \frac{1}{u ^{0.2}}

125 \cdot \frac{1}{u ^{0.2}} = \frac{125}{u ^{0.2}}

125 \cdot \frac{1}{u ^{0.2}}

分式相乘:

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

= \frac{1 \cdot 125}{u ^{0.2}}

数字相乘:

1 \cdot 125 = 125

= \frac{125}{u ^{0.2}}

= 125 u^{0.2} + \frac{125}{u ^{0.2}}

展开

75 (u^{0.4} + u^{- 0.4}) : 75 u^{0.4} + \frac{75}{u ^{0.4}}

75 (u^{0.4} + u^{- 0.4})

使用指数法则:

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

= 75 (u^{0.4} + \frac{1}{u ^{0.4}})

使用分配律:

a (b + c) = ab + a c

a = 75, b = u^{0.4}, c = \frac{1}{u ^{0.4}}

= 75 u^{0.4} + 75 \cdot \frac{1}{u ^{0.4}}

75 \cdot \frac{1}{u ^{0.4}} = \frac{75}{u ^{0.4}}

75 \cdot \frac{1}{u ^{0.4}}

分式相乘:

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

= \frac{1 \cdot 75}{u ^{0.4}}

数字相乘:

1 \cdot 75 = 75

= \frac{75}{u ^{0.4}}

= 75 u^{0.4} + \frac{75}{u ^{0.4}}

125 u^{0.2} + \frac{125}{u ^{0.2}} = 75 u^{0.4} + \frac{75}{u ^{0.4}}

使用以下的指数特性

: a^{n} = (m a)^{(n \cdot m)}

u^{0.4} = (5 u)^{(0.4 \cdot 5)}

125 5 u + \frac{125}{5 u} = 75 (5 u)^{2} + \frac{75}{( 5 u ) ^{2}}

用

5 u = v

改写方程式

125 v + \frac{125}{v} = 75 v^{2} + \frac{75}{v ^{2}}

解

125 v + \frac{125}{v} = 75 v^{2} + \frac{75}{v ^{2}} : v \approx 0.50859 \dots, v \approx 1.96621 \dots

125 v + \frac{125}{v} = 75 v^{2} + \frac{75}{v ^{2}}

乘以最小公倍数

125 v + \frac{125}{v} = 75 v^{2} + \frac{75}{v ^{2}}

找到

v, v^{2}

的最小公倍数:

v^{2}

v, v^{2}

最小公倍数 (LCM)

计算出由出现在

v

或

v^{2}

中的因子组成的表达式

= v^{2}

乘以最小公倍数=

v^{2}

125 v v^{2} + \frac{125}{v} v^{2} = 75 v^{2} v^{2} + \frac{75}{v ^{2}} v^{2}

化简

125 v v^{2} + \frac{125}{v} v^{2} = 75 v^{2} v^{2} + \frac{75}{v ^{2}} v^{2}

化简

125 v v^{2} : 125 v^{3}

125 v v^{2}

使用指数法则:

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

v v^{2} = v^{1 + 2}

= 125 v^{1 + 2}

数字相加:

1 + 2 = 3

= 125 v^{3}

化简

\frac{125}{v} v^{2} : 125 v

\frac{125}{v} v^{2}

分式相乘:

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

= \frac{125 v ^{2}}{v}

约分:

v

= 125 v

化简

75 v^{2} v^{2} : 75 v^{4}

75 v^{2} v^{2}

使用指数法则:

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

v^{2} v^{2} = v^{2 + 2}

= 75 v^{2 + 2}

数字相加:

2 + 2 = 4

= 75 v^{4}

化简

\frac{75}{v ^{2}} v^{2} : 75

\frac{75}{v ^{2}} v^{2}

分式相乘:

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

= \frac{75 v ^{2}}{v ^{2}}

约分:

v^{2}

= 75

125 v^{3} + 125 v = 75 v^{4} + 75

125 v^{3} + 125 v = 75 v^{4} + 75

125 v^{3} + 125 v = 75 v^{4} + 75

解

125 v^{3} + 125 v = 75 v^{4} + 75 : v \approx 0.50859 \dots, v \approx 1.96621 \dots

125 v^{3} + 125 v = 75 v^{4} + 75

交换两边

75 v^{4} + 75 = 125 v^{3} + 125 v

将

125 v

para o lado esquerdo

75 v^{4} + 75 = 125 v^{3} + 125 v

两边减去

125 v

75 v^{4} + 75 - 125 v = 125 v^{3} + 125 v - 125 v

化简

75 v^{4} + 75 - 125 v = 125 v^{3}

75 v^{4} + 75 - 125 v = 125 v^{3}

将

125 v^{3}

para o lado esquerdo

75 v^{4} + 75 - 125 v = 125 v^{3}

两边减去

125 v^{3}

75 v^{4} + 75 - 125 v - 125 v^{3} = 125 v^{3} - 125 v^{3}

化简

75 v^{4} + 75 - 125 v - 125 v^{3} = 0

75 v^{4} + 75 - 125 v - 125 v^{3} = 0

改写成标准形式

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

75 v^{4} - 125 v^{3} - 125 v + 75 = 0

使用牛顿-拉弗森方法找到

75 v^{4} - 125 v^{3} - 125 v + 75 = 0

的一个解:

v \approx 0.50859 \dots

75 v^{4} - 125 v^{3} - 125 v + 75 = 0

牛顿-拉弗森近似法定义

f (v) = 75 v^{4} - 125 v^{3} - 125 v + 75

找到

f^{^{'}} (v) : 300 v^{3} - 375 v^{2} - 125

\frac{d}{d v} (75 v^{4} - 125 v^{3} - 125 v + 75)

使用微分加减法定则:

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

= \frac{d}{d v} (75 v^{4}) - \frac{d}{d v} (125 v^{3}) - \frac{d}{d v} (125 v) + \frac{d}{d v} (75)

\frac{d}{d v} (75 v^{4}) = 300 v^{3}

\frac{d}{d v} (75 v^{4})

将常数提出:

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

= 75 \frac{d}{d v} (v^{4})

使用幂法则:

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

= 75 \cdot 4 v^{4 - 1}

化简

= 300 v^{3}

\frac{d}{d v} (125 v^{3}) = 375 v^{2}

\frac{d}{d v} (125 v^{3})

将常数提出:

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

= 125 \frac{d}{d v} (v^{3})

使用幂法则:

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

= 125 \cdot 3 v^{3 - 1}

化简

= 375 v^{2}

\frac{d}{d v} (125 v) = 125

\frac{d}{d v} (125 v)

将常数提出:

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

= 125 \frac{d v}{d v}

使用常见微分定则:

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

= 125 \cdot 1

化简

= 125

\frac{d}{d v} (75) = 0

\frac{d}{d v} (75)

常数微分:

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

= 0

= 300 v^{3} - 375 v^{2} - 125 + 0

化简

= 300 v^{3} - 375 v^{2} - 125

令

v_{0} = 1

计算

v_{n + 1}

至

Δ v_{n + 1} < 0.000001

v_{1} = 0.5 : Δ v_{1} = 0.5

f (v_{0}) = 75 \cdot 1^{4} - 125 \cdot 1^{3} - 125 \cdot 1 + 75 = - 100

f^{^{'}} (v_{0}) = 300 \cdot 1^{3} - 375 \cdot 1^{2} - 125 = - 200

v_{1} = 0.5

Δ v_{1} = ∣ 0.5 - 1 ∣ = 0.5

Δ v_{1} = 0.5

v_{2} = 0.50862 \dots : Δ v_{2} = 0.00862 \dots

f (v_{1}) = 75 \cdot 0. 5^{4} - 125 \cdot 0. 5^{3} - 125 \cdot 0.5 + 75 = 1.5625

f^{^{'}} (v_{1}) = 300 \cdot 0. 5^{3} - 375 \cdot 0. 5^{2} - 125 = - 181.25

v_{2} = 0.50862 \dots

Δ v_{2} = ∣ 0.50862 \dots - 0.5 ∣ = 0.00862 \dots

Δ v_{2} = 0.00862 \dots

v_{3} = 0.50859 \dots : Δ v_{3} = 0.00003 \dots

f (v_{2}) = 75 \cdot 0.50862 \dots^{4} - 125 \cdot 0.50862 \dots^{3} - 125 \cdot 0.50862 \dots + 75 = - 0.00555 \dots

f^{^{'}} (v_{2}) = 300 \cdot 0.50862 \dots^{3} - 375 \cdot 0.50862 \dots^{2} - 125 = - 182.53733 \dots

v_{3} = 0.50859 \dots

Δ v_{3} = ∣ 0.50859 \dots - 0.50862 \dots ∣ = 0.00003 \dots

Δ v_{3} = 0.00003 \dots

v_{4} = 0.50859 \dots : Δ v_{4} = 3.77392 E - 10

f (v_{3}) = 75 \cdot 0.50859 \dots^{4} - 125 \cdot 0.50859 \dots^{3} - 125 \cdot 0.50859 \dots + 75 = - 6.88864 E - 8

f^{^{'}} (v_{3}) = 300 \cdot 0.50859 \dots^{3} - 375 \cdot 0.50859 \dots^{2} - 125 = - 182.53281 \dots

v_{4} = 0.50859 \dots

Δ v_{4} = ∣ 0.50859 \dots - 0.50859 \dots ∣ = 3.77392 E - 10

Δ v_{4} = 3.77392 E - 10

v \approx 0.50859 \dots

使用长除法

Eq u a t i o n 0 : \frac{75 v ^{4} - 125 v ^{3} - 125 v + 75}{v - 0.50859 \dots} = 75 v^{3} - 86.85573 \dots v^{2} - 44.17397 \dots v - 147.46645 \dots

75 v^{3} - 86.85573 \dots v^{2} - 44.17397 \dots v - 147.46645 \dots \approx 0

使用牛顿-拉弗森方法找到

75 v^{3} - 86.85573 \dots v^{2} - 44.17397 \dots v - 147.46645 \dots = 0

的一个解:

v \approx 1.96621 \dots

75 v^{3} - 86.85573 \dots v^{2} - 44.17397 \dots v - 147.46645 \dots = 0

牛顿-拉弗森近似法定义

f (v) = 75 v^{3} - 86.85573 \dots v^{2} - 44.17397 \dots v - 147.46645 \dots

找到

f^{^{'}} (v) : 225 v^{2} - 173.71146 \dots v - 44.17397 \dots

\frac{d}{d v} (75 v^{3} - 86.85573 \dots v^{2} - 44.17397 \dots v - 147.46645 \dots)

使用微分加减法定则:

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

= \frac{d}{d v} (75 v^{3}) - \frac{d}{d v} (86.85573 \dots v^{2}) - \frac{d}{d v} (44.17397 \dots v) - \frac{d}{d v} (147.46645 \dots)

\frac{d}{d v} (75 v^{3}) = 225 v^{2}

\frac{d}{d v} (75 v^{3})

将常数提出:

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

= 75 \frac{d}{d v} (v^{3})

使用幂法则:

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

= 75 \cdot 3 v^{3 - 1}

化简

= 225 v^{2}

\frac{d}{d v} (86.85573 \dots v^{2}) = 173.71146 \dots v

\frac{d}{d v} (86.85573 \dots v^{2})

将常数提出:

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

= 86.85573 \dots \frac{d}{d v} (v^{2})

使用幂法则:

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

= 86.85573 \dots \cdot 2 v^{2 - 1}

化简

= 173.71146 \dots v

\frac{d}{d v} (44.17397 \dots v) = 44.17397 \dots

\frac{d}{d v} (44.17397 \dots v)

将常数提出:

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

= 44.17397 \dots \frac{d v}{d v}

使用常见微分定则:

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

= 44.17397 \dots \cdot 1

化简

= 44.17397 \dots

\frac{d}{d v} (147.46645 \dots) = 0

\frac{d}{d v} (147.46645 \dots)

常数微分:

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

= 0

= 225 v^{2} - 173.71146 \dots v - 44.17397 \dots - 0

化简

= 225 v^{2} - 173.71146 \dots v - 44.17397 \dots

令

v_{0} = - 3

计算

v_{n + 1}

至

Δ v_{n + 1} < 0.000001

v_{1} = - 1.87222 \dots : Δ v_{1} = 1.12777 \dots

f (v_{0}) = 75 (- 3)^{3} - 86.85573 \dots (- 3)^{2} - 44.17397 \dots (- 3) - 147.46645 \dots = - 2821.64610 \dots

f^{^{'}} (v_{0}) = 225 (- 3)^{2} - 173.71146 \dots (- 3) - 44.17397 \dots = 2501.96041 \dots

v_{1} = - 1.87222 \dots

Δ v_{1} = ∣ - 1.87222 \dots - (- 3) ∣ = 1.12777 \dots

Δ v_{1} = 1.12777 \dots

v_{2} = - 1.06697 \dots : Δ v_{2} = 0.80525 \dots

f (v_{1}) = 75 (- 1.87222 \dots)^{3} - 86.85573 \dots (- 1.87222 \dots)^{2} - 44.17397 \dots (- 1.87222 \dots) - 147.46645 \dots = - 861.40576 \dots

f^{^{'}} (v_{1}) = 225 (- 1.87222 \dots)^{2} - 173.71146 \dots (- 1.87222 \dots) - 44.17397 \dots = 1069.72985 \dots

v_{2} = - 1.06697 \dots

Δ v_{2} = ∣ - 1.06697 \dots - (- 1.87222 \dots) ∣ = 0.80525 \dots

Δ v_{2} = 0.80525 \dots

v_{3} = - 0.33628 \dots : Δ v_{3} = 0.73068 \dots

f (v_{2}) = 75 (- 1.06697 \dots)^{3} - 86.85573 \dots (- 1.06697 \dots)^{2} - 44.17397 \dots (- 1.06697 \dots) - 147.46645 \dots = - 290.31297 \dots

f^{^{'}} (v_{2}) = 225 (- 1.06697 \dots)^{2} - 173.71146 \dots (- 1.06697 \dots) - 44.17397 \dots = 397.31688 \dots

v_{3} = - 0.33628 \dots

Δ v_{3} = ∣ - 0.33628 \dots - (- 1.06697 \dots) ∣ = 0.73068 \dots

Δ v_{3} = 0.73068 \dots

v_{4} = 3.32442 \dots : Δ v_{4} = 3.66071 \dots

f (v_{3}) = 75 (- 0.33628 \dots)^{3} - 86.85573 \dots (- 0.33628 \dots)^{2} - 44.17397 \dots (- 0.33628 \dots) - 147.46645 \dots = - 145.28601 \dots

f^{^{'}} (v_{3}) = 225 (- 0.33628 \dots)^{2} - 173.71146 \dots (- 0.33628 \dots) - 44.17397 \dots = 39.68787 \dots

v_{4} = 3.32442 \dots

Δ v_{4} = ∣ 3.32442 \dots - (- 0.33628 \dots) ∣ = 3.66071 \dots

Δ v_{4} = 3.66071 \dots

v_{5} = 2.51941 \dots : Δ v_{5} = 0.80501 \dots

f (v_{4}) = 75 \cdot 3.32442 \dots^{3} - 86.85573 \dots \cdot 3.32442 \dots^{2} - 44.17397 \dots \cdot 3.32442 \dots - 147.46645 \dots = 1501.34112 \dots

f^{^{'}} (v_{4}) = 225 \cdot 3.32442 \dots^{2} - 173.71146 \dots \cdot 3.32442 \dots - 44.17397 \dots = 1864.99529 \dots

v_{5} = 2.51941 \dots

Δ v_{5} = ∣ 2.51941 \dots - 3.32442 \dots ∣ = 0.80501 \dots

Δ v_{5} = 0.80501 \dots

v_{6} = 2.10802 \dots : Δ v_{6} = 0.41139 \dots

f (v_{5}) = 75 \cdot 2.51941 \dots^{3} - 86.85573 \dots \cdot 2.51941 \dots^{2} - 44.17397 \dots \cdot 2.51941 \dots - 147.46645 \dots = 389.32098 \dots

f^{^{'}} (v_{5}) = 225 \cdot 2.51941 \dots^{2} - 173.71146 \dots \cdot 2.51941 \dots - 44.17397 \dots = 946.35416 \dots

v_{6} = 2.10802 \dots

Δ v_{6} = ∣ 2.10802 \dots - 2.51941 \dots ∣ = 0.41139 \dots

Δ v_{6} = 0.41139 \dots

v_{7} = 1.97907 \dots : Δ v_{7} = 0.12895 \dots

f (v_{6}) = 75 \cdot 2.10802 \dots^{3} - 86.85573 \dots \cdot 2.10802 \dots^{2} - 44.17397 \dots \cdot 2.10802 \dots - 147.46645 \dots = 76.01658 \dots

f^{^{'}} (v_{6}) = 225 \cdot 2.10802 \dots^{2} - 173.71146 \dots \cdot 2.10802 \dots - 44.17397 \dots = 589.48794 \dots

v_{7} = 1.97907 \dots

Δ v_{7} = ∣ 1.97907 \dots - 2.10802 \dots ∣ = 0.12895 \dots

Δ v_{7} = 0.12895 \dots

v_{8} = 1.96633 \dots : Δ v_{8} = 0.01273 \dots

f (v_{7}) = 75 \cdot 1.97907 \dots^{3} - 86.85573 \dots \cdot 1.97907 \dots^{2} - 44.17397 \dots \cdot 1.97907 \dots - 147.46645 \dots = 6.28209 \dots

f^{^{'}} (v_{7}) = 225 \cdot 1.97907 \dots^{2} - 173.71146 \dots \cdot 1.97907 \dots - 44.17397 \dots = 493.30322 \dots

v_{8} = 1.96633 \dots

Δ v_{8} = ∣ 1.96633 \dots - 1.97907 \dots ∣ = 0.01273 \dots

Δ v_{8} = 0.01273 \dots

v_{9} = 1.96621 \dots : Δ v_{9} = 0.00011 \dots

f (v_{8}) = 75 \cdot 1.96633 \dots^{3} - 86.85573 \dots \cdot 1.96633 \dots^{2} - 44.17397 \dots \cdot 1.96633 \dots - 147.46645 \dots = 0.05797 \dots

f^{^{'}} (v_{8}) = 225 \cdot 1.96633 \dots^{2} - 173.71146 \dots \cdot 1.96633 \dots - 44.17397 \dots = 484.21052 \dots

v_{9} = 1.96621 \dots

Δ v_{9} = ∣ 1.96621 \dots - 1.96633 \dots ∣ = 0.00011 \dots

Δ v_{9} = 0.00011 \dots

v_{10} = 1.96621 \dots : Δ v_{10} = 1.05283 E - 8

f (v_{9}) = 75 \cdot 1.96621 \dots^{3} - 86.85573 \dots \cdot 1.96621 \dots^{2} - 44.17397 \dots \cdot 1.96621 \dots - 147.46645 \dots = 5.097 E - 6

f^{^{'}} (v_{9}) = 225 \cdot 1.96621 \dots^{2} - 173.71146 \dots \cdot 1.96621 \dots - 44.17397 \dots = 484.12538 \dots

v_{10} = 1.96621 \dots

Δ v_{10} = ∣ 1.96621 \dots - 1.96621 \dots ∣ = 1.05283 E - 8

Δ v_{10} = 1.05283 E - 8

v \approx 1.96621 \dots

使用长除法

Eq u a t i o n 0 : \frac{75 v ^{3} - 86.85573 \dots v ^{2} - 44.17397 \dots v - 147.46645 \dots}{v - 1.96621 \dots} = 75 v^{2} + 60.61072 \dots v + 75

75 v^{2} + 60.61072 \dots v + 75 \approx 0

使用牛顿-拉弗森方法找到

75 v^{2} + 60.61072 \dots v + 75 = 0

的一个解:

v \in R

无解

75 v^{2} + 60.61072 \dots v + 75 = 0

牛顿-拉弗森近似法定义

f (v) = 75 v^{2} + 60.61072 \dots v + 75

找到

f^{^{'}} (v) : 150 v + 60.61072 \dots

\frac{d}{d v} (75 v^{2} + 60.61072 \dots v + 75)

使用微分加减法定则:

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

= \frac{d}{d v} (75 v^{2}) + \frac{d}{d v} (60.61072 \dots v) + \frac{d}{d v} (75)

\frac{d}{d v} (75 v^{2}) = 150 v

\frac{d}{d v} (75 v^{2})

将常数提出:

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

= 75 \frac{d}{d v} (v^{2})

使用幂法则:

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

= 75 \cdot 2 v^{2 - 1}

化简

= 150 v

\frac{d}{d v} (60.61072 \dots v) = 60.61072 \dots

\frac{d}{d v} (60.61072 \dots v)

将常数提出:

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

= 60.61072 \dots \frac{d v}{d v}

使用常见微分定则:

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

= 60.61072 \dots \cdot 1

化简

= 60.61072 \dots

\frac{d}{d v} (75) = 0

\frac{d}{d v} (75)

常数微分:

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

= 0

= 150 v + 60.61072 \dots + 0

化简

= 150 v + 60.61072 \dots

令

v_{0} = - 1

计算

v_{n + 1}

至

Δ v_{n + 1} < 0.000001

v_{1} = 6.66134 E - 16 : Δ v_{1} = 1

f (v_{0}) = 75 (- 1)^{2} + 60.61072 \dots (- 1) + 75 = 89.38927 \dots

f^{^{'}} (v_{0}) = 150 (- 1) + 60.61072 \dots = - 89.38927 \dots

v_{1} = 6.66134 E - 16

Δ v_{1} = ∣ 6.66134 E - 16 - (- 1) ∣ = 1

Δ v_{1} = 1

v_{2} = - 1.23740 \dots : Δ v_{2} = 1.23740 \dots

f (v_{1}) = 75 \cdot 6.66134 E - 1 6^{2} + 60.61072 \dots \cdot 6.66134 E - 16 + 75 = 75

f^{^{'}} (v_{1}) = 150 \cdot 6.66134 E - 16 + 60.61072 \dots = 60.61072 \dots

v_{2} = - 1.23740 \dots

Δ v_{2} = ∣ - 1.23740 \dots - 6.66134 E - 16 ∣ = 1.23740 \dots

Δ v_{2} = 1.23740 \dots

v_{3} = - 0.31870 \dots : Δ v_{3} = 0.91870 \dots

f (v_{2}) = 75 (- 1.23740 \dots)^{2} + 60.61072 \dots (- 1.23740 \dots) + 75 = 114.83780 \dots

f^{^{'}} (v_{2}) = 150 (- 1.23740 \dots) + 60.61072 \dots = - 124.99999 \dots

v_{3} = - 0.31870 \dots

Δ v_{3} = ∣ - 0.31870 \dots - (- 1.23740 \dots) ∣ = 0.91870 \dots

Δ v_{3} = 0.91870 \dots

v_{4} = - 5.26202 \dots : Δ v_{4} = 4.94332 \dots

f (v_{3}) = 75 (- 0.31870 \dots)^{2} + 60.61072 \dots (- 0.31870 \dots) + 75 = 63.30105 \dots

f^{^{'}} (v_{3}) = 150 (- 0.31870 \dots) + 60.61072 \dots = 12.80536 \dots

v_{4} = - 5.26202 \dots

Δ v_{4} = ∣ - 5.26202 \dots - (- 0.31870 \dots) ∣ = 4.94332 \dots

Δ v_{4} = 4.94332 \dots

v_{5} = - 2.74693 \dots : Δ v_{5} = 2.51509 \dots

f (v_{4}) = 75 (- 5.26202 \dots)^{2} + 60.61072 \dots (- 5.26202 \dots) + 75 = 1832.73443 \dots

f^{^{'}} (v_{4}) = 150 (- 5.26202 \dots) + 60.61072 \dots = - 728.69334 \dots

v_{5} = - 2.74693 \dots

Δ v_{5} = ∣ - 2.74693 \dots - (- 5.26202 \dots) ∣ = 2.51509 \dots

Δ v_{5} = 2.51509 \dots

v_{6} = - 1.39693 \dots : Δ v_{6} = 1.34999 \dots

f (v_{5}) = 75 (- 2.74693 \dots)^{2} + 60.61072 \dots (- 2.74693 \dots) + 75 = 474.42846 \dots

f^{^{'}} (v_{5}) = 150 (- 2.74693 \dots) + 60.61072 \dots = - 351.42879 \dots

v_{6} = - 1.39693 \dots

Δ v_{6} = ∣ - 1.39693 \dots - (- 2.74693 \dots) ∣ = 1.34999 \dots

Δ v_{6} = 1.34999 \dots

v_{7} = - 0.47912 \dots : Δ v_{7} = 0.91780 \dots

f (v_{6}) = 75 (- 1.39693 \dots)^{2} + 60.61072 \dots (- 1.39693 \dots) + 75 = 136.68726 \dots

f^{^{'}} (v_{6}) = 150 (- 1.39693 \dots) + 60.61072 \dots = - 148.92897 \dots

v_{7} = - 0.47912 \dots

Δ v_{7} = ∣ - 0.47912 \dots - (- 1.39693 \dots) ∣ = 0.91780 \dots

Δ v_{7} = 0.91780 \dots

v_{8} = 5.13225 \dots : Δ v_{8} = 5.61138 \dots

f (v_{7}) = 75 (- 0.47912 \dots)^{2} + 60.61072 \dots (- 0.47912 \dots) + 75 = 63.17699 \dots

f^{^{'}} (v_{7}) = 150 (- 0.47912 \dots) + 60.61072 \dots = - 11.25871 \dots

v_{8} = 5.13225 \dots

Δ v_{8} = ∣ 5.13225 \dots - (- 0.47912 \dots) ∣ = 5.61138 \dots

Δ v_{8} = 5.61138 \dots

v_{9} = 2.28852 \dots : Δ v_{9} = 2.84373 \dots

f (v_{8}) = 75 \cdot 5.13225 \dots^{2} + 60.61072 \dots \cdot 5.13225 \dots + 75 = 2361.57320 \dots

f^{^{'}} (v_{8}) = 150 \cdot 5.13225 \dots + 60.61072 \dots = 830.44904 \dots

v_{9} = 2.28852 \dots

Δ v_{9} = ∣ 2.28852 \dots - 5.13225 \dots ∣ = 2.84373 \dots

Δ v_{9} = 2.84373 \dots

v_{10} = 0.78685 \dots : Δ v_{10} = 1.50167 \dots

f (v_{9}) = 75 \cdot 2.28852 \dots^{2} + 60.61072 \dots \cdot 2.28852 \dots + 75 = 606.51019 \dots

f^{^{'}} (v_{9}) = 150 \cdot 2.28852 \dots + 60.61072 \dots = 403.88948 \dots

v_{10} = 0.78685 \dots

Δ v_{10} = ∣ 0.78685 \dots - 2.28852 \dots ∣ = 1.50167 \dots

Δ v_{10} = 1.50167 \dots

v_{11} = - 0.15990 \dots : Δ v_{11} = 0.94675 \dots

f (v_{10}) = 75 \cdot 0.78685 \dots^{2} + 60.61072 \dots \cdot 0.78685 \dots + 75 = 169.12677 \dots

f^{^{'}} (v_{10}) = 150 \cdot 0.78685 \dots + 60.61072 \dots = 178.63844 \dots

v_{11} = - 0.15990 \dots

Δ v_{11} = ∣ - 0.15990 \dots - 0.78685 \dots ∣ = 0.94675 \dots

Δ v_{11} = 0.94675 \dots

v_{12} = - 1.99540 \dots : Δ v_{12} = 1.83550 \dots

f (v_{11}) = 75 (- 0.15990 \dots)^{2} + 60.61072 \dots (- 0.15990 \dots) + 75 = 67.22582 \dots

f^{^{'}} (v_{11}) = 150 (- 0.15990 \dots) + 60.61072 \dots = 36.62524 \dots

v_{12} = - 1.99540 \dots

Δ v_{12} = ∣ - 1.99540 \dots - (- 0.15990 \dots) ∣ = 1.83550 \dots

Δ v_{12} = 1.83550 \dots

无法得出解

解为

v \approx 0.50859 \dots, v \approx 1.96621 \dots

v \approx 0.50859 \dots, v \approx 1.96621 \dots

验证解

找到无定义的点（奇点）:

v = 0

取

125 v + \frac{125}{v}

的分母，令其等于零

v = 0

取

75 v^{2} + \frac{75}{v ^{2}}

的分母，令其等于零

解

v^{2} = 0 : v = 0

v^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

v = 0

以下点无定义

v = 0

将不在定义域的点与解相综合:

v \approx 0.50859 \dots, v \approx 1.96621 \dots

v \approx 0.50859 \dots, v \approx 1.96621 \dots

代回

v = 5 u

，求解

u

解

5 u = 0.50859 \dots : u = 0.03402 \dots

5 u = 0.50859 \dots

对方程式两边

5

次方:

u = 0.03402 \dots

5 u = 0.50859 \dots

(5 u)^{5} = 0.50859 \dots^{5}

展开

(5 u)^{5} : u

(5 u)^{5}

使用根式运算法则:

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

= (u^{\frac{1}{5}})^{5}

使用指数法则:

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

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

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

\frac{1}{5} \cdot 5

分式相乘:

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

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

约分:

5

= 1

= u

展开

0.50859 \dots^{5} : 0.03402 \dots

0.50859 \dots^{5}

0.50859 \dots^{5} = 0.03402 \dots

= 0.03402 \dots

u = 0.03402 \dots

u = 0.03402 \dots

验证解:

u = 0.03402 \dots

真

将它们代入

5 u = 0.50859 \dots

检验解是否符合

去除与方程不符的解。

代入

u = 0.03402 \dots :

真

5 0.03402 \dots = 0.50859 \dots

5 0.03402 \dots = 0.50859 \dots

5 0.03402 \dots

5 0.03402 \dots = 0.50859 \dots

= 0.50859 \dots

0.50859 \dots = 0.50859 \dots

真

解是

u = 0.03402 \dots

解

5 u = 1.96621 \dots : u = 29.38731 \dots

5 u = 1.96621 \dots

对方程式两边

5

次方:

u = 29.38731 \dots

5 u = 1.96621 \dots

(5 u)^{5} = 1.96621 \dots^{5}

展开

(5 u)^{5} : u

(5 u)^{5}

使用根式运算法则:

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

= (u^{\frac{1}{5}})^{5}

使用指数法则:

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

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

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

\frac{1}{5} \cdot 5

分式相乘:

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

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

约分:

5

= 1

= u

展开

1.96621 \dots^{5} : 29.38731 \dots

1.96621 \dots^{5}

1.96621 \dots^{5} = 29.38731 \dots

= 29.38731 \dots

u = 29.38731 \dots

u = 29.38731 \dots

验证解:

u = 29.38731 \dots

真

将它们代入

5 u = 1.96621 \dots

检验解是否符合

去除与方程不符的解。

代入

u = 29.38731 \dots :

真

5 29.38731 \dots = 1.96621 \dots

5 29.38731 \dots = 1.96621 \dots

5 29.38731 \dots

5 29.38731 \dots = 1.96621 \dots

= 1.96621 \dots

1.96621 \dots = 1.96621 \dots

真

解是

u = 29.38731 \dots

u = 0.03402 \dots, u = 29.38731 \dots

验证解:

u = 0.03402 \dots

真

, u = 29.38731 \dots

真

将它们代入

125 (u^{0.2} + u^{- 0.2}) = 75 (u^{0.4} + u^{- 0.4})

检验解是否符合

去除与方程不符的解。

代入

u = 0.03402 \dots :

真

125 (0.03402 \dots^{0.2} + 0.03402 \dots^{- 0.2}) = 75 (0.03402 \dots^{0.4} + 0.03402 \dots^{- 0.4})

125 (0.03402 \dots^{0.2} + 0.03402 \dots^{- 0.2}) = 309.35120 \dots

125 (0.03402 \dots^{0.2} + 0.03402 \dots^{- 0.2})

0.03402 \dots^{0.2} = 0.50859 \dots

= 125 (0.50859 \dots + 0.03402 \dots^{- 0.2})

0.03402 \dots^{- 0.2} = 1.96621 \dots

= 125 (0.50859 \dots + 1.96621 \dots)

数字相加:

0.50859 \dots + 1.96621 \dots = 2.47480 \dots

= 125 \cdot 2.47480 \dots

数字相乘:

125 \cdot 2.47480 \dots = 309.35120 \dots

= 309.35120 \dots

75 (0.03402 \dots^{0.4} + 0.03402 \dots^{- 0.4}) = 309.35120 \dots

75 (0.03402 \dots^{0.4} + 0.03402 \dots^{- 0.4})

0.03402 \dots^{0.4} = 0.25866 \dots

= 75 (0.25866 \dots + 0.03402 \dots^{- 0.4})

0.03402 \dots^{- 0.4} = 3.86601 \dots

= 75 (0.25866 \dots + 3.86601 \dots)

数字相加:

0.25866 \dots + 3.86601 \dots = 4.12468 \dots

= 75 \cdot 4.12468 \dots

数字相乘:

75 \cdot 4.12468 \dots = 309.35120 \dots

= 309.35120 \dots

309.35120 \dots = 309.35120 \dots

真

代入

u = 29.38731 \dots :

真

125 (29.38731 \dots^{0.2} + 29.38731 \dots^{- 0.2}) = 75 (29.38731 \dots^{0.4} + 29.38731 \dots^{- 0.4})

125 (29.38731 \dots^{0.2} + 29.38731 \dots^{- 0.2}) = 309.35120 \dots

125 (29.38731 \dots^{0.2} + 29.38731 \dots^{- 0.2})

29.38731 \dots^{0.2} = 1.96621 \dots

= 125 (1.96621 \dots + 29.38731 \dots^{- 0.2})

29.38731 \dots^{- 0.2} = 0.50859 \dots

= 125 (0.50859 \dots + 1.96621 \dots)

数字相加:

1.96621 \dots + 0.50859 \dots = 2.47480 \dots

= 125 \cdot 2.47480 \dots

数字相乘:

125 \cdot 2.47480 \dots = 309.35120 \dots

= 309.35120 \dots

75 (29.38731 \dots^{0.4} + 29.38731 \dots^{- 0.4}) = 309.35120 \dots

75 (29.38731 \dots^{0.4} + 29.38731 \dots^{- 0.4})

29.38731 \dots^{0.4} = 3.86601 \dots

= 75 (3.86601 \dots + 29.38731 \dots^{- 0.4})

29.38731 \dots^{- 0.4} = 0.25866 \dots

= 75 (0.25866 \dots + 3.86601 \dots)

数字相加:

3.86601 \dots + 0.25866 \dots = 4.12468 \dots

= 75 \cdot 4.12468 \dots

数字相乘:

75 \cdot 4.12468 \dots = 309.35120 \dots

= 309.35120 \dots

309.35120 \dots = 309.35120 \dots

真

解为

u = 0.03402 \dots, u = 29.38731 \dots

u = 0.03402 \dots, u = 29.38731 \dots

代回

u = e^{x}

，求解

x

解

e^{x} = 0.03402 \dots : x = ln (0.03402 \dots)

e^{x} = 0.03402 \dots

使用指数运算法则

e^{x} = 0.03402 \dots

若

f (x) = g (x)

，则

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

ln (e^{x}) = ln (0.03402 \dots)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (0.03402 \dots)

x = ln (0.03402 \dots)

解

e^{x} = 29.38731 \dots : x = ln (29.38731 \dots)

e^{x} = 29.38731 \dots

使用指数运算法则

e^{x} = 29.38731 \dots

若

f (x) = g (x)

，则

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

ln (e^{x}) = ln (29.38731 \dots)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (29.38731 \dots)

x = ln (29.38731 \dots)

x = ln (0.03402 \dots), x = ln (29.38731 \dots)

验证解:

x = ln (0.03402 \dots)

真

, x = ln (29.38731 \dots)

真

将它们代入

\frac{\frac{e ^{0.2 x} + e ^{- 0.2 x}}{2}}{\frac{e ^{0.4 x} + e ^{- 0.4 x}}{2}} = \frac{75}{125}

检验解是否符合

去除与方程不符的解。

代入

x = ln (0.03402 \dots) :

真

\frac{\frac{e ^{0.2 l n (0.03402 \dots)} + e ^{- 0.2 l n (0.03402 \dots)}}{2}}{\frac{e ^{0.4 l n (0.03402 \dots)} + e ^{- 0.4 l n (0.03402 \dots)}}{2}} = \frac{75}{125}

\frac{\frac{e ^{0.2 l n (0.03402 \dots)} + e ^{- 0.2 l n (0.03402 \dots)}}{2}}{\frac{e ^{0.4 l n (0.03402 \dots)} + e ^{- 0.4 l n (0.03402 \dots)}}{2}} = 0.6

\frac{\frac{e ^{0.2 l n (0.03402 \dots)} + e ^{- 0.2 l n (0.03402 \dots)}}{2}}{\frac{e ^{0.4 l n (0.03402 \dots)} + e ^{- 0.4 l n (0.03402 \dots)}}{2}}

分式相除:

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

= \frac{( e ^{0.2 l n (0.03402 \dots)} + e ^{- 0.2 l n (0.03402 \dots)} ) \cdot 2}{2 ( e ^{0.4 l n (0.03402 \dots)} + e ^{- 0.4 l n (0.03402 \dots)} )}

约分:

2

= \frac{e ^{0.2 l n (0.03402 \dots)} + e ^{- 0.2 l n (0.03402 \dots)}}{e ^{0.4 l n (0.03402 \dots)} + e ^{- 0.4 l n (0.03402 \dots)}}

e^{0.4 l n (0.03402 \dots)} = 0.03402 \dots^{0.4}

e^{0.4 l n (0.03402 \dots)}

使用指数法则:

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

= (e^{l n (0.03402 \dots)})^{0.4}

使用对数计算法则:

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

e^{l n (0.03402 \dots)} = 0.03402 \dots

= 0.03402 \dots^{0.4}

e^{- 0.4 l n (0.03402 \dots)} = 0.03402 \dots^{- 0.4}

e^{- 0.4 l n (0.03402 \dots)}

使用指数法则:

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

= (e^{l n (0.03402 \dots)})^{- 0.4}

使用对数计算法则:

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

e^{l n (0.03402 \dots)} = 0.03402 \dots

= 0.03402 \dots^{- 0.4}

= \frac{e ^{0.2 l n (0.03402 \dots)} + e ^{- 0.2 l n (0.03402 \dots)}}{0.03402 \dots ^{0.4} + 0.03402 \dots ^{- 0.4}}

e^{0.2 l n (0.03402 \dots)} = 0.03402 \dots^{0.2}

e^{0.2 l n (0.03402 \dots)}

使用指数法则:

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

= (e^{l n (0.03402 \dots)})^{0.2}

使用对数计算法则:

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

e^{l n (0.03402 \dots)} = 0.03402 \dots

= 0.03402 \dots^{0.2}

e^{- 0.2 l n (0.03402 \dots)} = 0.03402 \dots^{- 0.2}

e^{- 0.2 l n (0.03402 \dots)}

使用指数法则:

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

= (e^{l n (0.03402 \dots)})^{- 0.2}

使用对数计算法则:

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

e^{l n (0.03402 \dots)} = 0.03402 \dots

= 0.03402 \dots^{- 0.2}

= \frac{0.03402 \dots ^{0.2} + 0.03402 \dots ^{- 0.2}}{0.03402 \dots ^{0.4} + 0.03402 \dots ^{- 0.4}}

化简

\frac{0.03402 \dots ^{0.2} + 0.03402 \dots ^{- 0.2}}{0.03402 \dots ^{0.4} + 0.03402 \dots ^{- 0.4}}

使用指数法则:

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

0.03402 \dots^{- 0.4} = \frac{1}{0.03402 \dots ^{0.4}}

= \frac{0.03402 \dots ^{0.2} + 0.03402 \dots ^{- 0.2}}{0.03402 \dots ^{0.4} + \frac{1}{0.03402 \dots ^{0.4}}}

使用指数法则:

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

0.03402 \dots^{- 0.2} = \frac{1}{0.03402 \dots ^{0.2}}

= \frac{0.03402 \dots ^{0.2} + \frac{1}{0.03402 \dots ^{0.2}}}{0.03402 \dots ^{0.4} + \frac{1}{0.03402 \dots ^{0.4}}}

化简

0.03402 \dots^{0.4} + \frac{1}{0.03402 \dots ^{0.4}} : 4.12468 \dots

0.03402 \dots^{0.4} + \frac{1}{0.03402 \dots ^{0.4}}

将项转换为分式:

0.03402 \dots^{0.4} = \frac{0.03402 \dots ^{0.4} \cdot 0.03402 \dots ^{0.4}}{0.03402 \dots ^{0.4}}

= \frac{0.03402 \dots ^{0.4} \cdot 0.03402 \dots ^{0.4}}{0.03402 \dots ^{0.4}} + \frac{1}{0.03402 \dots ^{0.4}}

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{0.03402 \dots ^{0.4} \cdot 0.03402 \dots ^{0.4} + 1}{0.03402 \dots ^{0.4}}

0.03402 \dots^{0.4} \cdot 0.03402 \dots^{0.4} + 1 = 0.03402 \dots^{0.8} + 1

0.03402 \dots^{0.4} \cdot 0.03402 \dots^{0.4} + 1

0.03402 \dots^{0.4} \cdot 0.03402 \dots^{0.4} = 0.03402 \dots^{0.8}

0.03402 \dots^{0.4} \cdot 0.03402 \dots^{0.4}

使用指数法则:

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

0.03402 \dots^{0.4} \cdot 0.03402 \dots^{0.4} = 0.03402 \dots^{0.4 + 0.4}

= 0.03402 \dots^{0.4 + 0.4}

数字相加:

0.4 + 0.4 = 0.8

= 0.03402 \dots^{0.8}

= 0.03402 \dots^{0.8} + 1

= \frac{0.03402 \dots ^{0.8} + 1}{0.03402 \dots ^{0.4}}

0.03402 \dots^{0.8} = 0.06690 \dots

= \frac{0.06690 \dots + 1}{0.03402 \dots ^{0.4}}

数字相加:

0.06690 \dots + 1 = 1.06690 \dots

= \frac{1.06690 \dots}{0.03402 \dots ^{0.4}}

0.03402 \dots^{0.4} = 0.25866 \dots

= \frac{1.06690 \dots}{0.25866 \dots}

数字相除:

\frac{1.06690 \dots}{0.25866 \dots} = 4.12468 \dots

= 4.12468 \dots

= \frac{0.03402 \dots ^{0.2} + \frac{1}{0.03402 \dots ^{0.2}}}{4.12468 \dots}

化简

0.03402 \dots^{0.2} + \frac{1}{0.03402 \dots ^{0.2}} : 2.47480 \dots

0.03402 \dots^{0.2} + \frac{1}{0.03402 \dots ^{0.2}}

将项转换为分式:

0.03402 \dots^{0.2} = \frac{0.03402 \dots ^{0.2} \cdot 0.03402 \dots ^{0.2}}{0.03402 \dots ^{0.2}}

= \frac{0.03402 \dots ^{0.2} \cdot 0.03402 \dots ^{0.2}}{0.03402 \dots ^{0.2}} + \frac{1}{0.03402 \dots ^{0.2}}

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{0.03402 \dots ^{0.2} \cdot 0.03402 \dots ^{0.2} + 1}{0.03402 \dots ^{0.2}}

0.03402 \dots^{0.2} \cdot 0.03402 \dots^{0.2} + 1 = 0.03402 \dots^{0.4} + 1

0.03402 \dots^{0.2} \cdot 0.03402 \dots^{0.2} + 1

0.03402 \dots^{0.2} \cdot 0.03402 \dots^{0.2} = 0.03402 \dots^{0.4}

0.03402 \dots^{0.2} \cdot 0.03402 \dots^{0.2}

使用指数法则:

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

0.03402 \dots^{0.2} \cdot 0.03402 \dots^{0.2} = 0.03402 \dots^{0.2 + 0.2}

= 0.03402 \dots^{0.2 + 0.2}

数字相加:

0.2 + 0.2 = 0.4

= 0.03402 \dots^{0.4}

= 0.03402 \dots^{0.4} + 1

= \frac{0.03402 \dots ^{0.4} + 1}{0.03402 \dots ^{0.2}}

0.03402 \dots^{0.4} = 0.25866 \dots

= \frac{0.25866 \dots + 1}{0.03402 \dots ^{0.2}}

数字相加:

0.25866 \dots + 1 = 1.25866 \dots

= \frac{1.25866 \dots}{0.03402 \dots ^{0.2}}

0.03402 \dots^{0.2} = 0.50859 \dots

= \frac{1.25866 \dots}{0.50859 \dots}

数字相除:

\frac{1.25866 \dots}{0.50859 \dots} = 2.47480 \dots

= 2.47480 \dots

= \frac{2.47480 \dots}{4.12468 \dots}

数字相除:

\frac{2.47480 \dots}{4.12468 \dots} = 0.6

= 0.6

= 0.6

\frac{75}{125} = \frac{3}{5}

\frac{75}{125}

约分:

25

= \frac{3}{5}

0.6 = \frac{3}{5}

真

代入

x = ln (29.38731 \dots) :

真

\frac{\frac{e ^{0.2 l n (29.38731 \dots)} + e ^{- 0.2 l n (29.38731 \dots)}}{2}}{\frac{e ^{0.4 l n (29.38731 \dots)} + e ^{- 0.4 l n (29.38731 \dots)}}{2}} = \frac{75}{125}

\frac{\frac{e ^{0.2 l n (29.38731 \dots)} + e ^{- 0.2 l n (29.38731 \dots)}}{2}}{\frac{e ^{0.4 l n (29.38731 \dots)} + e ^{- 0.4 l n (29.38731 \dots)}}{2}} = 0.6

\frac{\frac{e ^{0.2 l n (29.38731 \dots)} + e ^{- 0.2 l n (29.38731 \dots)}}{2}}{\frac{e ^{0.4 l n (29.38731 \dots)} + e ^{- 0.4 l n (29.38731 \dots)}}{2}}

分式相除:

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

= \frac{( e ^{0.2 l n (29.38731 \dots)} + e ^{- 0.2 l n (29.38731 \dots)} ) \cdot 2}{2 ( e ^{0.4 l n (29.38731 \dots)} + e ^{- 0.4 l n (29.38731 \dots)} )}

约分:

2

= \frac{e ^{0.2 l n (29.38731 \dots)} + e ^{- 0.2 l n (29.38731 \dots)}}{e ^{0.4 l n (29.38731 \dots)} + e ^{- 0.4 l n (29.38731 \dots)}}

e^{0.4 l n (29.38731 \dots)} = 29.38731 \dots^{0.4}

e^{0.4 l n (29.38731 \dots)}

使用指数法则:

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

= (e^{l n (29.38731 \dots)})^{0.4}

使用对数计算法则:

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

e^{l n (29.38731 \dots)} = 29.38731 \dots

= 29.38731 \dots^{0.4}

e^{- 0.4 l n (29.38731 \dots)} = 29.38731 \dots^{- 0.4}

e^{- 0.4 l n (29.38731 \dots)}

使用指数法则:

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

= (e^{l n (29.38731 \dots)})^{- 0.4}

使用对数计算法则:

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

e^{l n (29.38731 \dots)} = 29.38731 \dots

= 29.38731 \dots^{- 0.4}

= \frac{e ^{0.2 l n (29.38731 \dots)} + e ^{- 0.2 l n (29.38731 \dots)}}{29.38731 \dots ^{0.4} + 29.38731 \dots ^{- 0.4}}

e^{0.2 l n (29.38731 \dots)} = 29.38731 \dots^{0.2}

e^{0.2 l n (29.38731 \dots)}

使用指数法则:

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

= (e^{l n (29.38731 \dots)})^{0.2}

使用对数计算法则:

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

e^{l n (29.38731 \dots)} = 29.38731 \dots

= 29.38731 \dots^{0.2}

e^{- 0.2 l n (29.38731 \dots)} = 29.38731 \dots^{- 0.2}

e^{- 0.2 l n (29.38731 \dots)}

使用指数法则:

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

= (e^{l n (29.38731 \dots)})^{- 0.2}

使用对数计算法则:

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

e^{l n (29.38731 \dots)} = 29.38731 \dots

= 29.38731 \dots^{- 0.2}

= \frac{29.38731 \dots ^{0.2} + 29.38731 \dots ^{- 0.2}}{29.38731 \dots ^{0.4} + 29.38731 \dots ^{- 0.4}}

化简

\frac{29.38731 \dots ^{0.2} + 29.38731 \dots ^{- 0.2}}{29.38731 \dots ^{0.4} + 29.38731 \dots ^{- 0.4}}

使用指数法则:

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

29.38731 \dots^{- 0.4} = \frac{1}{29.38731 \dots ^{0.4}}

= \frac{29.38731 \dots ^{0.2} + 29.38731 \dots ^{- 0.2}}{29.38731 \dots ^{0.4} + \frac{1}{29.38731 \dots ^{0.4}}}

使用指数法则:

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

29.38731 \dots^{- 0.2} = \frac{1}{29.38731 \dots ^{0.2}}

= \frac{29.38731 \dots ^{0.2} + \frac{1}{29.38731 \dots ^{0.2}}}{29.38731 \dots ^{0.4} + \frac{1}{29.38731 \dots ^{0.4}}}

化简

29.38731 \dots^{0.4} + \frac{1}{29.38731 \dots ^{0.4}} : 4.12468 \dots

29.38731 \dots^{0.4} + \frac{1}{29.38731 \dots ^{0.4}}

将项转换为分式:

29.38731 \dots^{0.4} = \frac{29.38731 \dots ^{0.4} \cdot 29.38731 \dots ^{0.4}}{29.38731 \dots ^{0.4}}

= \frac{29.38731 \dots ^{0.4} \cdot 29.38731 \dots ^{0.4}}{29.38731 \dots ^{0.4}} + \frac{1}{29.38731 \dots ^{0.4}}

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{29.38731 \dots ^{0.4} \cdot 29.38731 \dots ^{0.4} + 1}{29.38731 \dots ^{0.4}}

29.38731 \dots^{0.4} \cdot 29.38731 \dots^{0.4} + 1 = 29.38731 \dots^{0.8} + 1

29.38731 \dots^{0.4} \cdot 29.38731 \dots^{0.4} + 1

29.38731 \dots^{0.4} \cdot 29.38731 \dots^{0.4} = 29.38731 \dots^{0.8}

29.38731 \dots^{0.4} \cdot 29.38731 \dots^{0.4}

使用指数法则:

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

29.38731 \dots^{0.4} \cdot 29.38731 \dots^{0.4} = 29.38731 \dots^{0.4 + 0.4}

= 29.38731 \dots^{0.4 + 0.4}

数字相加:

0.4 + 0.4 = 0.8

= 29.38731 \dots^{0.8}

= 29.38731 \dots^{0.8} + 1

= \frac{29.38731 \dots ^{0.8} + 1}{29.38731 \dots ^{0.4}}

29.38731 \dots^{0.8} = 14.94610 \dots

= \frac{14.94610 \dots + 1}{29.38731 \dots ^{0.4}}

数字相加:

14.94610 \dots + 1 = 15.94610 \dots

= \frac{15.94610 \dots}{29.38731 \dots ^{0.4}}

29.38731 \dots^{0.4} = 3.86601 \dots

= \frac{15.94610 \dots}{3.86601 \dots}

数字相除:

\frac{15.94610 \dots}{3.86601 \dots} = 4.12468 \dots

= 4.12468 \dots

= \frac{29.38731 \dots ^{0.2} + \frac{1}{29.38731 \dots ^{0.2}}}{4.12468 \dots}

化简

29.38731 \dots^{0.2} + \frac{1}{29.38731 \dots ^{0.2}} : 2.47480 \dots

29.38731 \dots^{0.2} + \frac{1}{29.38731 \dots ^{0.2}}

将项转换为分式:

29.38731 \dots^{0.2} = \frac{29.38731 \dots ^{0.2} \cdot 29.38731 \dots ^{0.2}}{29.38731 \dots ^{0.2}}

= \frac{29.38731 \dots ^{0.2} \cdot 29.38731 \dots ^{0.2}}{29.38731 \dots ^{0.2}} + \frac{1}{29.38731 \dots ^{0.2}}

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{29.38731 \dots ^{0.2} \cdot 29.38731 \dots ^{0.2} + 1}{29.38731 \dots ^{0.2}}

29.38731 \dots^{0.2} \cdot 29.38731 \dots^{0.2} + 1 = 29.38731 \dots^{0.4} + 1

29.38731 \dots^{0.2} \cdot 29.38731 \dots^{0.2} + 1

29.38731 \dots^{0.2} \cdot 29.38731 \dots^{0.2} = 29.38731 \dots^{0.4}

29.38731 \dots^{0.2} \cdot 29.38731 \dots^{0.2}

使用指数法则:

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

29.38731 \dots^{0.2} \cdot 29.38731 \dots^{0.2} = 29.38731 \dots^{0.2 + 0.2}

= 29.38731 \dots^{0.2 + 0.2}

数字相加:

0.2 + 0.2 = 0.4

= 29.38731 \dots^{0.4}

= 29.38731 \dots^{0.4} + 1

= \frac{29.38731 \dots ^{0.4} + 1}{29.38731 \dots ^{0.2}}

29.38731 \dots^{0.4} = 3.86601 \dots

= \frac{3.86601 \dots + 1}{29.38731 \dots ^{0.2}}

数字相加:

3.86601 \dots + 1 = 4.86601 \dots

= \frac{4.86601 \dots}{29.38731 \dots ^{0.2}}

29.38731 \dots^{0.2} = 1.96621 \dots

= \frac{4.86601 \dots}{1.96621 \dots}

数字相除:

\frac{4.86601 \dots}{1.96621 \dots} = 2.47480 \dots

= 2.47480 \dots

= \frac{2.47480 \dots}{4.12468 \dots}

数字相除:

\frac{2.47480 \dots}{4.12468 \dots} = 0.6

= 0.6

= 0.6

\frac{75}{125} = \frac{3}{5}

\frac{75}{125}

约分:

25

= \frac{3}{5}

0.6 = \frac{3}{5}

真

解为

x = ln (0.03402 \dots), x = ln (29.38731 \dots)

x = ln (0.03402 \dots), x = ln (29.38731 \dots)

作图

流行的例子

3cos(x)+6=cos(x)+5

3 cos (x) + 6 = cos (x) + 5

cos(x)= 1/3 ,(3pi)/2 <x<2pi

cos (x) = \frac{1}{3}, \frac{3 π}{2} < x < 2 π

(2.68)/(sin(126))=(1.2)/(sin(x))

\frac{2.68}{sin ( 12 6 ^{\circ} )} = \frac{1.2}{sin ( x )}

3csc^2(x)-5csc(x)-2=0

3 csc^{2} (x) - 5 csc (x) - 2 = 0

((sin(5x)))/((-1cos(5x)))=0

\frac{( sin ( 5 x ) )}{( - 1 cos ( 5 x ) )} = 0