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受欢迎的三角函数 >

2cosh^2(x)-5sinh(x)=5

解答

2 cosh^{2} (x) - 5 sinh (x) = 5

解答

x = ln (0.61803 \dots), x = ln (6.16227 \dots)

+1

度数

x = - 27.57140 \dots^{\circ}, x = 104.18930 \dots^{\circ}

求解步骤

2 cosh^{2} (x) - 5 sinh (x) = 5

使用三角恒等式改写

2 cosh^{2} (x) - 5 sinh (x) = 5

使用双曲函数恒等式:

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

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

使用双曲函数恒等式:

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

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

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

2 (\frac{e ^{x} + e ^{- x}}{2})^{2} - 5 \cdot \frac{e ^{x} - e ^{- x}}{2} = 5 : x = ln (0.61803 \dots), x = ln (6.16227 \dots)

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

使用指数运算法则

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

使用指数法则:

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

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

2 (\frac{e ^{x} + ( e ^{x} ) ^{- 1}}{2})^{2} - 5 \cdot \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{2} = 5

2 (\frac{e ^{x} + ( e ^{x} ) ^{- 1}}{2})^{2} - 5 \cdot \frac{e ^{x} - ( e ^{x} ) ^{- 1}}{2} = 5

用

e^{x} = u

改写方程式

2 (\frac{u + ( u ) ^{- 1}}{2})^{2} - 5 \cdot \frac{u - ( u ) ^{- 1}}{2} = 5

解

2 (\frac{u + u ^{- 1}}{2})^{2} - 5 \cdot \frac{u - u ^{- 1}}{2} = 5 : u \approx - 0.16227 \dots, u \approx 0.61803 \dots, u \approx - 1.61803 \dots, u \approx 6.16227 \dots

2 (\frac{u + u ^{- 1}}{2})^{2} - 5 \cdot \frac{u - u ^{- 1}}{2} = 5

整理后得

\frac{( u ^{2} + 1 ) ^{2}}{2 u ^{2}} - \frac{5 ( u ^{2} - 1 )}{2 u} = 5

乘以最小公倍数

\frac{( u ^{2} + 1 ) ^{2}}{2 u ^{2}} - \frac{5 ( u ^{2} - 1 )}{2 u} = 5

找到

2 u^{2}, 2 u

的最小公倍数:

2 u^{2}

2 u^{2}, 2 u

最小公倍数 (LCM)

2, 2

的最小公倍数:

2

2, 2

最小公倍数 (LCM)

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

将每个因子乘以它在

2

或

2

中出现的最多次数

= 2

数字相乘:

2 = 2

= 2

计算出由出现在

2 u^{2}

或

2 u

中的因子组成的表达式

= 2 u^{2}

乘以最小公倍数=

2 u^{2}

\frac{( u ^{2} + 1 ) ^{2}}{2 u ^{2}} \cdot 2 u^{2} - \frac{5 ( u ^{2} - 1 )}{2 u} \cdot 2 u^{2} = 5 \cdot 2 u^{2}

化简

\frac{( u ^{2} + 1 ) ^{2}}{2 u ^{2}} \cdot 2 u^{2} - \frac{5 ( u ^{2} - 1 )}{2 u} \cdot 2 u^{2} = 5 \cdot 2 u^{2}

化简

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

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

分式相乘:

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

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

约分:

2

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

约分:

u^{2}

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

化简

- \frac{5 ( u ^{2} - 1 )}{2 u} \cdot 2 u^{2} : - 5 u (u^{2} - 1)

- \frac{5 ( u ^{2} - 1 )}{2 u} \cdot 2 u^{2}

分式相乘:

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

= - \frac{5 ( u ^{2} - 1 ) \cdot 2 u ^{2}}{2 u}

约分:

2

= - \frac{5 ( u ^{2} - 1 ) u ^{2}}{u}

约分:

u

= - 5 u (u^{2} - 1)

化简

5 \cdot 2 u^{2} : 10 u^{2}

5 \cdot 2 u^{2}

数字相乘:

5 \cdot 2 = 10

= 10 u^{2}

(u^{2} + 1)^{2} - 5 u (u^{2} - 1) = 10 u^{2}

(u^{2} + 1)^{2} - 5 u (u^{2} - 1) = 10 u^{2}

(u^{2} + 1)^{2} - 5 u (u^{2} - 1) = 10 u^{2}

解

(u^{2} + 1)^{2} - 5 u (u^{2} - 1) = 10 u^{2} : u \approx - 0.16227 \dots, u \approx 0.61803 \dots, u \approx - 1.61803 \dots, u \approx 6.16227 \dots

(u^{2} + 1)^{2} - 5 u (u^{2} - 1) = 10 u^{2}

展开

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

(u^{2} + 1)^{2} - 5 u (u^{2} - 1)

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = - 5 u, b = u^{2}, c = 1

= - 5 u u^{2} - (- 5 u) \cdot 1

使用加减运算法则

- (- a) = a

= - 5 u^{2} u + 5 \cdot 1 \cdot u

化简

- 5 u^{2} u + 5 \cdot 1 \cdot u : - 5 u^{3} + 5 u

- 5 u^{2} u + 5 \cdot 1 \cdot u

5 u^{2} u = 5 u^{3}

5 u^{2} u

使用指数法则:

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

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

= 5 u^{2 + 1}

数字相加:

2 + 1 = 3

= 5 u^{3}

5 \cdot 1 \cdot u = 5 u

5 \cdot 1 \cdot u

数字相乘:

5 \cdot 1 = 5

= 5 u

= - 5 u^{3} + 5 u

= - 5 u^{3} + 5 u

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

u^{4} + 2 u^{2} + 1 - 5 u^{3} + 5 u = 10 u^{2}

将

10 u^{2}

para o lado esquerdo

u^{4} + 2 u^{2} + 1 - 5 u^{3} + 5 u = 10 u^{2}

两边减去

10 u^{2}

u^{4} + 2 u^{2} + 1 - 5 u^{3} + 5 u - 10 u^{2} = 10 u^{2} - 10 u^{2}

化简

u^{4} - 5 u^{3} - 8 u^{2} + 5 u + 1 = 0

u^{4} - 5 u^{3} - 8 u^{2} + 5 u + 1 = 0

使用牛顿-拉弗森方法找到

u^{4} - 5 u^{3} - 8 u^{2} + 5 u + 1 = 0

的一个解:

u \approx - 0.16227 \dots

u^{4} - 5 u^{3} - 8 u^{2} + 5 u + 1 = 0

牛顿-拉弗森近似法定义

f (u) = u^{4} - 5 u^{3} - 8 u^{2} + 5 u + 1

找到

f^{^{'}} (u) : 4 u^{3} - 15 u^{2} - 16 u + 5

\frac{d}{d u} (u^{4} - 5 u^{3} - 8 u^{2} + 5 u + 1)

使用微分加减法定则:

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

= \frac{d}{d u} (u^{4}) - \frac{d}{d u} (5 u^{3}) - \frac{d}{d u} (8 u^{2}) + \frac{d}{d u} (5 u) + \frac{d}{d u} (1)

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

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

使用幂法则:

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

= 4 u^{4 - 1}

化简

= 4 u^{3}

\frac{d}{d u} (5 u^{3}) = 15 u^{2}

\frac{d}{d u} (5 u^{3})

将常数提出:

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

= 5 \frac{d}{d u} (u^{3})

使用幂法则:

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

= 5 \cdot 3 u^{3 - 1}

化简

= 15 u^{2}

\frac{d}{d u} (8 u^{2}) = 16 u

\frac{d}{d u} (8 u^{2})

将常数提出:

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

= 8 \frac{d}{d u} (u^{2})

使用幂法则:

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

= 8 \cdot 2 u^{2 - 1}

化简

= 16 u

\frac{d}{d u} (5 u) = 5

\frac{d}{d u} (5 u)

将常数提出:

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

= 5 \frac{d u}{d u}

使用常见微分定则:

\frac{d u}{d u} = 1

= 5 \cdot 1

化简

= 5

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

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

常数微分:

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

= 0

= 4 u^{3} - 15 u^{2} - 16 u + 5 + 0

化简

= 4 u^{3} - 15 u^{2} - 16 u + 5

令

u_{0} = 0

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 0.2 : Δ u_{1} = 0.2

f (u_{0}) = 0^{4} - 5 \cdot 0^{3} - 8 \cdot 0^{2} + 5 \cdot 0 + 1 = 1

f^{^{'}} (u_{0}) = 4 \cdot 0^{3} - 15 \cdot 0^{2} - 16 \cdot 0 + 5 = 5

u_{1} = - 0.2

Δ u_{1} = ∣ - 0.2 - 0 ∣ = 0.2

Δ u_{1} = 0.2

u_{2} = - 0.16321 \dots : Δ u_{2} = 0.03678 \dots

f (u_{1}) = (- 0.2)^{4} - 5 (- 0.2)^{3} - 8 (- 0.2)^{2} + 5 (- 0.2) + 1 = - 0.2784

f^{^{'}} (u_{1}) = 4 (- 0.2)^{3} - 15 (- 0.2)^{2} - 16 (- 0.2) + 5 = 7.568

u_{2} = - 0.16321 \dots

Δ u_{2} = ∣ - 0.16321 \dots - (- 0.2) ∣ = 0.03678 \dots

Δ u_{2} = 0.03678 \dots

u_{3} = - 0.16227 \dots : Δ u_{3} = 0.00093 \dots

f (u_{2}) = (- 0.16321 \dots)^{4} - 5 (- 0.16321 \dots)^{3} - 8 (- 0.16321 \dots)^{2} + 5 (- 0.16321 \dots) + 1 = - 0.00672 \dots

f^{^{'}} (u_{2}) = 4 (- 0.16321 \dots)^{3} - 15 (- 0.16321 \dots)^{2} - 16 (- 0.16321 \dots) + 5 = 7.19444 \dots

u_{3} = - 0.16227 \dots

Δ u_{3} = ∣ - 0.16227 \dots - (- 0.16321 \dots) ∣ = 0.00093 \dots

Δ u_{3} = 0.00093 \dots

u_{4} = - 0.16227 \dots : Δ u_{4} = 6.57063 E - 7

f (u_{3}) = (- 0.16227 \dots)^{4} - 5 (- 0.16227 \dots)^{3} - 8 (- 0.16227 \dots)^{2} + 5 (- 0.16227 \dots) + 1 = - 4.72057 E - 6

f^{^{'}} (u_{3}) = 4 (- 0.16227 \dots)^{3} - 15 (- 0.16227 \dots)^{2} - 16 (- 0.16227 \dots) + 5 = 7.18434 \dots

u_{4} = - 0.16227 \dots

Δ u_{4} = ∣ - 0.16227 \dots - (- 0.16227 \dots) ∣ = 6.57063 E - 7

Δ u_{4} = 6.57063 E - 7

u \approx - 0.16227 \dots

使用长除法

Eq u a t i o n 0 : \frac{u ^{4} - 5 u ^{3} - 8 u ^{2} + 5 u + 1}{u + 0.16227 \dots} = u^{3} - 5.16227 \dots u^{2} - 7.16227 \dots u + 6.16227 \dots

u^{3} - 5.16227 \dots u^{2} - 7.16227 \dots u + 6.16227 \dots \approx 0

使用牛顿-拉弗森方法找到

u^{3} - 5.16227 \dots u^{2} - 7.16227 \dots u + 6.16227 \dots = 0

的一个解:

u \approx 0.61803 \dots

u^{3} - 5.16227 \dots u^{2} - 7.16227 \dots u + 6.16227 \dots = 0

牛顿-拉弗森近似法定义

f (u) = u^{3} - 5.16227 \dots u^{2} - 7.16227 \dots u + 6.16227 \dots

找到

f^{^{'}} (u) : 3 u^{2} - 10.32455 \dots u - 7.16227 \dots

\frac{d}{d u} (u^{3} - 5.16227 \dots u^{2} - 7.16227 \dots u + 6.16227 \dots)

使用微分加减法定则:

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

= \frac{d}{d u} (u^{3}) - \frac{d}{d u} (5.16227 \dots u^{2}) - \frac{d}{d u} (7.16227 \dots u) + \frac{d}{d u} (6.16227 \dots)

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

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

使用幂法则:

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

= 3 u^{3 - 1}

化简

= 3 u^{2}

\frac{d}{d u} (5.16227 \dots u^{2}) = 10.32455 \dots u

\frac{d}{d u} (5.16227 \dots u^{2})

将常数提出:

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

= 5.16227 \dots \frac{d}{d u} (u^{2})

使用幂法则:

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

= 5.16227 \dots \cdot 2 u^{2 - 1}

化简

= 10.32455 \dots u

\frac{d}{d u} (7.16227 \dots u) = 7.16227 \dots

\frac{d}{d u} (7.16227 \dots u)

将常数提出:

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

= 7.16227 \dots \frac{d u}{d u}

使用常见微分定则:

\frac{d u}{d u} = 1

= 7.16227 \dots \cdot 1

化简

= 7.16227 \dots

\frac{d}{d u} (6.16227 \dots) = 0

\frac{d}{d u} (6.16227 \dots)

常数微分:

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

= 0

= 3 u^{2} - 10.32455 \dots u - 7.16227 \dots + 0

化简

= 3 u^{2} - 10.32455 \dots u - 7.16227 \dots

令

u_{0} = 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 0.64365 \dots : Δ u_{1} = 0.35634 \dots

f (u_{0}) = 1^{3} - 5.16227 \dots \cdot 1^{2} - 7.16227 \dots \cdot 1 + 6.16227 \dots = - 5.16227 \dots

f^{^{'}} (u_{0}) = 3 \cdot 1^{2} - 10.32455 \dots \cdot 1 - 7.16227 \dots = - 14.48683 \dots

u_{1} = 0.64365 \dots

Δ u_{1} = ∣ 0.64365 \dots - 1 ∣ = 0.35634 \dots

Δ u_{1} = 0.35634 \dots

u_{2} = 0.61820 \dots : Δ u_{2} = 0.02545 \dots

f (u_{1}) = 0.64365 \dots^{3} - 5.16227 \dots \cdot 0.64365 \dots^{2} - 7.16227 \dots \cdot 0.64365 \dots + 6.16227 \dots = - 0.31981 \dots

f^{^{'}} (u_{1}) = 3 \cdot 0.64365 \dots^{2} - 10.32455 \dots \cdot 0.64365 \dots - 7.16227 \dots = - 12.56486 \dots

u_{2} = 0.61820 \dots

Δ u_{2} = ∣ 0.61820 \dots - 0.64365 \dots ∣ = 0.02545 \dots

Δ u_{2} = 0.02545 \dots

u_{3} = 0.61803 \dots : Δ u_{3} = 0.00017 \dots

f (u_{2}) = 0.61820 \dots^{3} - 5.16227 \dots \cdot 0.61820 \dots^{2} - 7.16227 \dots \cdot 0.61820 \dots + 6.16227 \dots = - 0.00210 \dots

f^{^{'}} (u_{2}) = 3 \cdot 0.61820 \dots^{2} - 10.32455 \dots \cdot 0.61820 \dots - 7.16227 \dots = - 12.39843 \dots

u_{3} = 0.61803 \dots

Δ u_{3} = ∣ 0.61803 \dots - 0.61820 \dots ∣ = 0.00017 \dots

Δ u_{3} = 0.00017 \dots

u_{4} = 0.61803 \dots : Δ u_{4} = 7.7271 E - 9

f (u_{3}) = 0.61803 \dots^{3} - 5.16227 \dots \cdot 0.61803 \dots^{2} - 7.16227 \dots \cdot 0.61803 \dots + 6.16227 \dots = - 9.57952 E - 8

f^{^{'}} (u_{3}) = 3 \cdot 0.61803 \dots^{2} - 10.32455 \dots \cdot 0.61803 \dots - 7.16227 \dots = - 12.39730 \dots

u_{4} = 0.61803 \dots

Δ u_{4} = ∣ 0.61803 \dots - 0.61803 \dots ∣ = 7.7271 E - 9

Δ u_{4} = 7.7271 E - 9

u \approx 0.61803 \dots

使用长除法

Eq u a t i o n 0 : \frac{u ^{3} - 5.16227 \dots u ^{2} - 7.16227 \dots u + 6.16227 \dots}{u - 0.61803 \dots} = u^{2} - 4.54424 \dots u - 9.97077 \dots

u^{2} - 4.54424 \dots u - 9.97077 \dots \approx 0

使用牛顿-拉弗森方法找到

u^{2} - 4.54424 \dots u - 9.97077 \dots = 0

的一个解:

u \approx - 1.61803 \dots

u^{2} - 4.54424 \dots u - 9.97077 \dots = 0

牛顿-拉弗森近似法定义

f (u) = u^{2} - 4.54424 \dots u - 9.97077 \dots

找到

f^{^{'}} (u) : 2 u - 4.54424 \dots

\frac{d}{d u} (u^{2} - 4.54424 \dots u - 9.97077 \dots)

使用微分加减法定则:

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

= \frac{d}{d u} (u^{2}) - \frac{d}{d u} (4.54424 \dots u) - \frac{d}{d u} (9.97077 \dots)

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

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

使用幂法则:

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

= 2 u^{2 - 1}

化简

= 2 u

\frac{d}{d u} (4.54424 \dots u) = 4.54424 \dots

\frac{d}{d u} (4.54424 \dots u)

将常数提出:

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

= 4.54424 \dots \frac{d u}{d u}

使用常见微分定则:

\frac{d u}{d u} = 1

= 4.54424 \dots \cdot 1

化简

= 4.54424 \dots

\frac{d}{d u} (9.97077 \dots) = 0

\frac{d}{d u} (9.97077 \dots)

常数微分:

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

= 0

= 2 u - 4.54424 \dots - 0

化简

= 2 u - 4.54424 \dots

令

u_{0} = - 2

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 1.63510 \dots : Δ u_{1} = 0.36489 \dots

f (u_{0}) = (- 2)^{2} - 4.54424 \dots (- 2) - 9.97077 \dots = 3.11771 \dots

f^{^{'}} (u_{0}) = 2 (- 2) - 4.54424 \dots = - 8.54424 \dots

u_{1} = - 1.63510 \dots

Δ u_{1} = ∣ - 1.63510 \dots - (- 2) ∣ = 0.36489 \dots

Δ u_{1} = 0.36489 \dots

u_{2} = - 1.61807 \dots : Δ u_{2} = 0.01703 \dots

f (u_{1}) = (- 1.63510 \dots)^{2} - 4.54424 \dots (- 1.63510 \dots) - 9.97077 \dots = 0.13314 \dots

f^{^{'}} (u_{1}) = 2 (- 1.63510 \dots) - 4.54424 \dots = - 7.81446 \dots

u_{2} = - 1.61807 \dots

Δ u_{2} = ∣ - 1.61807 \dots - (- 1.63510 \dots) ∣ = 0.01703 \dots

Δ u_{2} = 0.01703 \dots

u_{3} = - 1.61803 \dots : Δ u_{3} = 0.00003 \dots

f (u_{2}) = (- 1.61807 \dots)^{2} - 4.54424 \dots (- 1.61807 \dots) - 9.97077 \dots = 0.00029 \dots

f^{^{'}} (u_{2}) = 2 (- 1.61807 \dots) - 4.54424 \dots = - 7.78038 \dots

u_{3} = - 1.61803 \dots

Δ u_{3} = ∣ - 1.61803 \dots - (- 1.61807 \dots) ∣ = 0.00003 \dots

Δ u_{3} = 0.00003 \dots

u_{4} = - 1.61803 \dots : Δ u_{4} = 1.78938 E - 10

f (u_{3}) = (- 1.61803 \dots)^{2} - 4.54424 \dots (- 1.61803 \dots) - 9.97077 \dots = 1.3922 E - 9

f^{^{'}} (u_{3}) = 2 (- 1.61803 \dots) - 4.54424 \dots = - 7.78031 \dots

u_{4} = - 1.61803 \dots

Δ u_{4} = ∣ - 1.61803 \dots - (- 1.61803 \dots) ∣ = 1.78938 E - 10

Δ u_{4} = 1.78938 E - 10

u \approx - 1.61803 \dots

使用长除法

Eq u a t i o n 0 : \frac{u ^{2} - 4.54424 \dots u - 9.97077 \dots}{u + 1.61803 \dots} = u - 6.16227 \dots

u - 6.16227 \dots \approx 0

u \approx 6.16227 \dots

解为

u \approx - 0.16227 \dots, u \approx 0.61803 \dots, u \approx - 1.61803 \dots, u \approx 6.16227 \dots

u \approx - 0.16227 \dots, u \approx 0.61803 \dots, u \approx - 1.61803 \dots, u \approx 6.16227 \dots

验证解

找到无定义的点（奇点）:

u = 0

取

2 (\frac{u + u ^{- 1}}{2})^{2} - 5 \frac{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 \approx - 0.16227 \dots, u \approx 0.61803 \dots, u \approx - 1.61803 \dots, u \approx 6.16227 \dots

u \approx - 0.16227 \dots, u \approx 0.61803 \dots, u \approx - 1.61803 \dots, u \approx 6.16227 \dots

代回

u = e^{x}

，求解

x

解

e^{x} = - 0.16227 \dots : x \in R

无解

e^{x} = - 0.16227 \dots

a^{f (x)}

对于

x

不能为零或负值

\in R

x \in R 无解

解

e^{x} = 0.61803 \dots : x = ln (0.61803 \dots)

e^{x} = 0.61803 \dots

使用指数运算法则

e^{x} = 0.61803 \dots

若

f (x) = g (x)

，则

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

ln (e^{x}) = ln (0.61803 \dots)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (0.61803 \dots)

x = ln (0.61803 \dots)

解

e^{x} = - 1.61803 \dots : x \in R

无解

e^{x} = - 1.61803 \dots

a^{f (x)}

对于

x

不能为零或负值

\in R

x \in R 无解

解

e^{x} = 6.16227 \dots : x = ln (6.16227 \dots)

e^{x} = 6.16227 \dots

使用指数运算法则

e^{x} = 6.16227 \dots

若

f (x) = g (x)

，则

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

ln (e^{x}) = ln (6.16227 \dots)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (6.16227 \dots)

x = ln (6.16227 \dots)

x = ln (0.61803 \dots), x = ln (6.16227 \dots)

x = ln (0.61803 \dots), x = ln (6.16227 \dots)

作图

流行的例子

10+5cos(2x)=15cos(x)

10 + 5 cos (2 x) = 15 cos (x)

solvefor y,x=cos(2y)

so l v e f or y, x = cos (2 y)

tan (x) = \frac{a}{b}

solvefor x,arcsin(3x-4y)= pi/2

so l v e f or x, arcsin (3 x - 4 y) = \frac{π}{2}

sin (x) a = \frac{π}{3}