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受欢迎的三角函数 >

(1+tanh(x))/(1-tanh(x))=2

解答

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

解答

x = \frac{1}{2} ln (2)

+1

度数

x = 19.85720 \dots^{\circ}

求解步骤

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

使用三角恒等式改写

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

使用双曲函数恒等式:

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

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

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

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

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

在两边乘以

1 - \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}

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

化简

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

使用指数运算法则

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

使用指数法则:

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

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

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

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

用

e^{x} = u

改写方程式

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

解

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

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

整理后得

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

在两边乘以

u^{2} + 1

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

在两边乘以

u^{2} + 1

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

化简

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

化简

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

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

乘以:

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

= (u^{2} + 1)

去除括号:

(a) = a

= u^{2} + 1

化简

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

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

分式相乘:

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

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

约分:

u^{2} + 1

= u^{2} - 1

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

化简

u^{2} + 1 + u^{2} - 1 : 2 u^{2}

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

对同类项分组

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

同类项相加:

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

= 2 u^{2} + 1 - 1

1 - 1 = 0

= 2 u^{2}

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

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

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

展开

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

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

乘开

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

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

使用 FOIL 方法:

(a + b) (c + d) = a c + a d + b c + b d

a = 1, b = - \frac{u ^{2} - 1}{u ^{2} + 1}, c = u^{2}, d = 1

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

使用加减运算法则

+ (- a) = - a

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

化简

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

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

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

1 \cdot u^{2}

乘以:

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

= u^{2}

1 \cdot 1 = 1

1 \cdot 1

数字相乘:

1 \cdot 1 = 1

= 1

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

\frac{u ^{2} - 1}{u ^{2} + 1} u^{2}

分式相乘:

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

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

乘开

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

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

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

使用分配律:

a (b - c) = ab - a c

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

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

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

化简

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

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

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

= u^{2 + 2}

数字相加:

2 + 2 = 4

= u^{4}

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

1 \cdot u^{2}

乘以:

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

= u^{2}

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

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

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

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

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

乘以:

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

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

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

合并分式

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

使用法则

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

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

分解

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

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

分解

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

u^{4} - u^{2}

因式分解出通项

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

u^{4} - u^{2}

使用指数法则:

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

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

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

因式分解出通项

u^{2}

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

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

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

因式分解出通项

(u^{2} - 1)

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

分解

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

- u^{2} - 1

因式分解出通项

- 1

= - (u^{2} + 1)

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

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

约分:

u^{2} + 1

= - (u^{2} - 1)

取负

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

= - u^{2} + 1

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

对同类项分组

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

同类项相加:

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

= 1 + 1

数字相加:

1 + 1 = 2

= 2

= 2

= 2 \cdot 2

乘开

2 \cdot 2 : 4

2 \cdot 2

打开括号

= 2 \cdot 2

数字相乘:

2 \cdot 2 = 4

= 4

= 4

2 u^{2} = 4

解

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

2 u^{2} = 4

两边除以

2

2 u^{2} = 4

两边除以

2

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

化简

u^{2} = 2

u^{2} = 2

对于

x^{2} = f (a)

解为

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

u = 2, u = - 2

u = 2, u = - 2

验证解

找到无定义的点（奇点）:

u = 0

取

1 + \frac{u - u ^{- 1}}{u + u ^{- 1}}

的分母，令其等于零

u = 0

取

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

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 = 2, u = - 2

u = 2, u = - 2

代回

u = e^{x}

，求解

x

解

e^{x} = 2 : x = \frac{1}{2} ln (2)

e^{x} = 2

使用指数运算法则

e^{x} = 2

使用指数法则:

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

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

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

若

f (x) = g (x)

，则

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

ln (e^{x}) = ln (2^{\frac{1}{2}})

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

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

使用对数计算法则:

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

ln (2^{\frac{1}{2}}) = \frac{1}{2} ln (2)

x = \frac{1}{2} ln (2)

x = \frac{1}{2} ln (2)

解

e^{x} = - 2 : x \in R

无解

e^{x} = - 2

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 (2)

验证解:

x = \frac{1}{2} ln (2)

真

将它们代入

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

检验解是否符合

去除与方程不符的解。

代入

x = \frac{1}{2} ln (2) :

真

\frac{1 + \frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}}}{1 - \frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}}} = 2

\frac{1 + \frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}}}{1 - \frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}}} = 2

\frac{1 + \frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}}}{1 - \frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}}}

\frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}} = \frac{1}{3}

\frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}}

e^{\frac{1}{2} l n (2)} = 2

e^{\frac{1}{2} l n (2)}

使用指数法则:

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

= e^{l n (2)}

使用对数计算法则:

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

e^{l n (2)} = 2

= 2

e^{- \frac{1}{2} l n (2)} = \frac{1}{2}

e^{- \frac{1}{2} l n (2)}

使用指数法则:

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

= (e^{l n (2)})^{- \frac{1}{2}}

使用对数计算法则:

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

e^{l n (2)} = 2

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

使用指数法则:

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

= \frac{1}{2}

= \frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{2 + \frac{1}{2}}

e^{\frac{1}{2} l n (2)} = 2

e^{\frac{1}{2} l n (2)}

使用指数法则:

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

= e^{l n (2)}

使用对数计算法则:

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

e^{l n (2)} = 2

= 2

e^{- \frac{1}{2} l n (2)} = \frac{1}{2}

e^{- \frac{1}{2} l n (2)}

使用指数法则:

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

= (e^{l n (2)})^{- \frac{1}{2}}

使用对数计算法则:

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

e^{l n (2)} = 2

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

使用指数法则:

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

= \frac{1}{2}

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

化简

2 + \frac{1}{2} : \frac{3}{2}

2 + \frac{1}{2}

将项转换为分式:

2 = \frac{2 2}{2}

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

22 + 1 = 3

22 + 1

使用根式运算法则:

a a = a

22 = 2

= 2 + 1

数字相加:

2 + 1 = 3

= 3

= \frac{3}{2}

= \frac{2 - \frac{1}{2}}{\frac{3}{2}}

化简

2 - \frac{1}{2} : \frac{1}{2}

2 - \frac{1}{2}

将项转换为分式:

2 = \frac{2 2}{2}

= \frac{2 2}{2} - \frac{1}{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{2 2 - 1}{2}

22 - 1 = 1

22 - 1

使用根式运算法则:

a a = a

22 = 2

= 2 - 1

数字相减:

2 - 1 = 1

= 1

= \frac{1}{2}

= \frac{\frac{1}{2}}{\frac{3}{2}}

分式相除:

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

= \frac{1 \cdot 2}{2 \cdot 3}

整理后得

= \frac{2}{2 \cdot 3}

约分:

2

= \frac{1}{3}

= \frac{1 + \frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}}}{1 - \frac{1}{3}}

\frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}} = \frac{1}{3}

\frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{e ^{\frac{1}{2} l n (2)} + e ^{- \frac{1}{2} l n (2)}}

e^{\frac{1}{2} l n (2)} = 2

e^{\frac{1}{2} l n (2)}

使用指数法则:

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

= e^{l n (2)}

使用对数计算法则:

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

e^{l n (2)} = 2

= 2

e^{- \frac{1}{2} l n (2)} = \frac{1}{2}

e^{- \frac{1}{2} l n (2)}

使用指数法则:

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

= (e^{l n (2)})^{- \frac{1}{2}}

使用对数计算法则:

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

e^{l n (2)} = 2

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

使用指数法则:

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

= \frac{1}{2}

= \frac{e ^{\frac{1}{2} l n (2)} - e ^{- \frac{1}{2} l n (2)}}{2 + \frac{1}{2}}

e^{\frac{1}{2} l n (2)} = 2

e^{\frac{1}{2} l n (2)}

使用指数法则:

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

= e^{l n (2)}

使用对数计算法则:

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

e^{l n (2)} = 2

= 2

e^{- \frac{1}{2} l n (2)} = \frac{1}{2}

e^{- \frac{1}{2} l n (2)}

使用指数法则:

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

= (e^{l n (2)})^{- \frac{1}{2}}

使用对数计算法则:

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

e^{l n (2)} = 2

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

使用指数法则:

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

= \frac{1}{2}

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

化简

2 + \frac{1}{2} : \frac{3}{2}

2 + \frac{1}{2}

将项转换为分式:

2 = \frac{2 2}{2}

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

22 + 1 = 3

22 + 1

使用根式运算法则:

a a = a

22 = 2

= 2 + 1

数字相加:

2 + 1 = 3

= 3

= \frac{3}{2}

= \frac{2 - \frac{1}{2}}{\frac{3}{2}}

化简

2 - \frac{1}{2} : \frac{1}{2}

2 - \frac{1}{2}

将项转换为分式:

2 = \frac{2 2}{2}

= \frac{2 2}{2} - \frac{1}{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{2 2 - 1}{2}

22 - 1 = 1

22 - 1

使用根式运算法则:

a a = a

22 = 2

= 2 - 1

数字相减:

2 - 1 = 1

= 1

= \frac{1}{2}

= \frac{\frac{1}{2}}{\frac{3}{2}}

分式相除:

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

= \frac{1 \cdot 2}{2 \cdot 3}

整理后得

= \frac{2}{2 \cdot 3}

约分:

2

= \frac{1}{3}

= \frac{1 + \frac{1}{3}}{1 - \frac{1}{3}}

化简

\frac{1 + \frac{1}{3}}{1 - \frac{1}{3}}

化简

1 - \frac{1}{3} : \frac{2}{3}

1 - \frac{1}{3}

将项转换为分式:

1 = \frac{1 \cdot 3}{3}

= \frac{1 \cdot 3}{3} - \frac{1}{3}

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{1 \cdot 3 - 1}{3}

1 \cdot 3 - 1 = 2

1 \cdot 3 - 1

数字相乘:

1 \cdot 3 = 3

= 3 - 1

数字相减:

3 - 1 = 2

= 2

= \frac{2}{3}

= \frac{1 + \frac{1}{3}}{\frac{2}{3}}

化简

1 + \frac{1}{3} : \frac{4}{3}

1 + \frac{1}{3}

将项转换为分式:

1 = \frac{1 \cdot 3}{3}

= \frac{1 \cdot 3}{3} + \frac{1}{3}

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{1 \cdot 3 + 1}{3}

1 \cdot 3 + 1 = 4

1 \cdot 3 + 1

数字相乘:

1 \cdot 3 = 3

= 3 + 1

数字相加:

3 + 1 = 4

= 4

= \frac{4}{3}

= \frac{\frac{4}{3}}{\frac{2}{3}}

分式相除:

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

= \frac{4 \cdot 3}{3 \cdot 2}

约分:

3

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

= 2

2 = 2

真

解是

x = \frac{1}{2} ln (2)

x = \frac{1}{2} ln (2)

作图

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