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solvefor u,x=sinh(u)

解答

求解

u, x = sinh (u)

解答

u = ln (x + x^{2} + 1), u = ln (x - x^{2} + 1)

求解步骤

x = sinh (u)

交换两边

sinh (u) = x

使用三角恒等式改写

sinh (u) = x

使用双曲函数恒等式:

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

\frac{e ^{u} - e ^{- u}}{2} = x

\frac{e ^{u} - e ^{- u}}{2} = x

\frac{e ^{u} - e ^{- u}}{2} = x : u = ln (x + x^{2} + 1), u = ln (x - x^{2} + 1)

\frac{e ^{u} - e ^{- u}}{2} = x

在两边乘以

2

\frac{e ^{u} - e ^{- u}}{2} \cdot 2 = x \cdot 2

化简

e^{u} - e^{- u} = x \cdot 2

使用指数运算法则

e^{u} - e^{- u} = x \cdot 2

使用指数法则:

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

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

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

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

用

e^{u} = v

改写方程式

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

解

v - v^{- 1} = x \cdot 2 : v = x + x^{2} + 1, v = x - x^{2} + 1

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

整理后得

v - \frac{1}{v} = x \cdot 2

在两边乘以

v

v - \frac{1}{v} = x \cdot 2

在两边乘以

v

vv - \frac{1}{v} v = x \cdot 2 v

化简

vv - \frac{1}{v} v = x \cdot 2 v

化简

vv : v^{2}

vv

使用指数法则:

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

vv = v^{1 + 1}

= v^{1 + 1}

数字相加:

1 + 1 = 2

= v^{2}

化简

- \frac{1}{v} v : - 1

- \frac{1}{v} v

分式相乘:

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

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

约分:

v

= - 1

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

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

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

解

v^{2} - 1 = x \cdot 2 v : v = x + x^{2} + 1, v = x - x^{2} + 1

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

将

x 2 v

para o lado esquerdo

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

两边减去

x 2 v

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

化简

v^{2} - 1 - x \cdot 2 v = 0

v^{2} - 1 - x \cdot 2 v = 0

改写成标准形式

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

v^{2} - x \cdot 2 v - 1 = 0

使用求根公式求解

v^{2} - x \cdot 2 v - 1 = 0

二次方程求根公式:

若

a = 1, b = - x 2, c = - 1

v_{1, 2} = \frac{- ( - x \cdot 2 ) \pm ( - x \cdot 2 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

v_{1, 2} = \frac{- ( - x \cdot 2 ) \pm ( - x \cdot 2 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

化简

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

(- x \cdot 2)^{2} - 4 \cdot 1 \cdot (- 1)

使用法则

- (- a) = a

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

(- x \cdot 2)^{2} = 2^{2} x^{2}

(- x \cdot 2)^{2}

使用指数法则:

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

若

n

是偶数

(- 2 x)^{2} = (2 x)^{2}

= (2 x)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} x^{2}

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4

= 2^{2} x^{2} + 4

分解

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

x^{2} \cdot 2^{2} + 4

改写为

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

因式分解出通项

4

= 4 (x^{2} + 1)

= 4 (x^{2} + 1)

使用根式运算法则:

假定

a \geq 0, b \geq 0

= 4 x^{2} + 1

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

2^{2} = 2

= 2

= 2 x^{2} + 1

v_{1, 2} = \frac{- ( - x \cdot 2 ) \pm 2 x ^{2} + 1}{2 \cdot 1}

将解分隔开

v_{1} = \frac{- ( - x \cdot 2 ) + 2 x ^{2} + 1}{2 \cdot 1}, v_{2} = \frac{- ( - x \cdot 2 ) - 2 x ^{2} + 1}{2 \cdot 1}

v = \frac{- ( - x 2 ) + 2 x ^{2} + 1}{2 \cdot 1} : x + x^{2} + 1

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

使用法则

- (- a) = a

= \frac{x \cdot 2 + 2 x ^{2} + 1}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{2 x + 2 x ^{2} + 1}{2}

因式分解出通项

2

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

数字相除:

\frac{2}{2} = 1

= x + x^{2} + 1

v = \frac{- ( - x 2 ) - 2 x ^{2} + 1}{2 \cdot 1} : x - x^{2} + 1

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

使用法则

- (- a) = a

= \frac{x \cdot 2 - 2 x ^{2} + 1}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{2 x - 2 x ^{2} + 1}{2}

因式分解出通项

2

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

数字相除:

\frac{2}{2} = 1

= x - x^{2} + 1

二次方程组的解是:

v = x + x^{2} + 1, v = x - x^{2} + 1

v = x + x^{2} + 1, v = x - x^{2} + 1

v = x + x^{2} + 1, v = x - x^{2} + 1

代回

v = e^{u}

，求解

u

解

e^{u} = x + x^{2} + 1 : u = ln (x + x^{2} + 1)

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

使用指数运算法则

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

若

f (x) = g (x)

，则

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

ln (e^{u}) = ln (x + x^{2} + 1)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{u}) = u

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

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

解

e^{u} = x - x^{2} + 1 : u = ln (x - x^{2} + 1)

e^{u} = x - x^{2} + 1

使用指数运算法则

e^{u} = x - x^{2} + 1

若

f (x) = g (x)

，则

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

ln (e^{u}) = ln (x - x^{2} + 1)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{u}) = u

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

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

u = ln (x + x^{2} + 1), u = ln (x - x^{2} + 1)

u = ln (x + x^{2} + 1), u = ln (x - x^{2} + 1)

流行的例子

sin(θ)=(tan(θ))/(sin(θ))

sin (θ) = \frac{tan ( θ )}{sin ( θ )}

5tan^2(t)-tan(t)=0

5 tan^{2} (t) - tan (t) = 0

sin (x) = \frac{9}{16}

24=32+8-2*sqrt(32*8)*cos(θ)

24 = 32 + 8 - 2 \cdot 32 \cdot 8 \cdot cos (θ)

sin(45+a)cos(45+a)=(sqrt(2))/2

sin (4 5^{\circ} + a) cos (4 5^{\circ} + a) = \frac{2}{2}