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受欢迎的三角函数 >

sqrt(2/(pi0.1)tan((pi0.1)/2))

解答

\frac{2}{π \cdot 0.1} tan (\frac{π \cdot 0.1}{2})

解答

\frac{2 5 π 4 - 3 2 5 + 5 - 10 5 + 5 + 11 + 4 5}{π}

+1

十进制

1.00414 \dots

求解步骤

\frac{2}{π 0.1} tan (\frac{π 0.1}{2})

= \frac{2}{π \frac{1}{10}} tan (\frac{π \frac{1}{10}}{2})

化简:

\frac{π \frac{1}{10}}{2} = \frac{π}{20}

\frac{π \frac{1}{10}}{2}

乘

π \frac{1}{10} : \frac{π}{10}

π \frac{1}{10}

分式相乘:

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

= \frac{1 π}{10}

乘以:

1 π = π

= \frac{π}{10}

= \frac{\frac{π}{10}}{2}

使用分式法则:

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

= \frac{π}{10 \cdot 2}

数字相乘:

10 \cdot 2 = 20

= \frac{π}{20}

\frac{2}{π \frac{1}{10}} tan (\frac{π}{20}) = \frac{2 5 π tan ( \frac{π}{20} )}{π}

\frac{2}{π \frac{1}{10}} tan (\frac{π}{20})

乘

\frac{2}{π \frac{1}{10}} tan (\frac{π}{20}) : \frac{20 tan ( \frac{π}{20} )}{π}

\frac{2}{π \frac{1}{10}} tan (\frac{π}{20})

分式相乘:

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

= \frac{2 tan ( \frac{π}{20} )}{π \frac{1}{10}}

乘

π \frac{1}{10} : \frac{π}{10}

π \frac{1}{10}

分式相乘:

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

= \frac{1 π}{10}

乘以:

1 π = π

= \frac{π}{10}

= \frac{2 tan ( \frac{π}{20} )}{\frac{π}{10}}

使用分式法则:

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

= \frac{2 tan ( \frac{π}{20} ) \cdot 10}{π}

数字相乘:

2 \cdot 10 = 20

= \frac{20 tan ( \frac{π}{20} )}{π}

= \frac{20 tan ( \frac{π}{20} )}{π}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{20 tan ( \frac{π}{20} )}{π}

使用根式运算法则:

n ab = n a n b,

假定

a \geq 0, b \geq 0

20 tan (\frac{π}{20}) = 20 tan (\frac{π}{20})

= \frac{20 tan ( \frac{π}{20} )}{π}

20 = 25

20

20

质因数分解:

2^{2} \cdot 5

20

20

除以

220 = 10 \cdot 2

= 2 \cdot 10

10

除以

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

使用根式运算法则:

n ab = n a n b

= 5 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 25

= \frac{2 5 tan ( \frac{π}{20} )}{π}

\frac{2 5 tan ( \frac{π}{20} )}{π}

有理化:

\frac{2 5 π tan ( \frac{π}{20} )}{π}

\frac{2 5 tan ( \frac{π}{20} )}{π}

乘以共轭根式

\frac{π}{π}

= \frac{2 5 tan ( \frac{π}{20} ) π}{π π}

π π = π

π π

使用根式运算法则:

a a = a

π π = π

= π

= \frac{2 5 π tan ( \frac{π}{20} )}{π}

= \frac{2 5 π tan ( \frac{π}{20} )}{π}

= \frac{2 5 π tan ( \frac{π}{20} )}{π}

使用三角恒等式改写:

tan (\frac{π}{20}) = - 32 5 + 5 - 10 5 + 5 + 11 + 45

tan (\frac{π}{20})

使用三角恒等式改写:

\frac{1 - cos ( \frac{π}{10} )}{1 + cos ( \frac{π}{10} )}

tan (\frac{π}{20})

将

tan (\frac{π}{20})

写为

tan (\frac{\frac{π}{10}}{2})

= tan (\frac{\frac{π}{10}}{2})

使用半角公式:

tan (\frac{θ}{2}) = \frac{1 - cos ( θ )}{1 + cos ( θ )}

使用三角恒等式改写:

tan^{2} (θ) = \frac{1 - cos ( 2 θ )}{1 + cos ( 2 θ )}

利用以下特性

tan (θ) = \frac{sin ( θ )}{cos ( θ )}

两边进行平方

tan^{2} (θ) = \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

使用三角恒等式改写:

sin^{2} (θ) = \frac{1 - cos ( 2 θ )}{2}

使用倍角公式

cos (2 θ) = 1 - 2 sin^{2} (θ)

交换两边

2 sin^{2} (θ) - 1 = - cos (2 θ)

两边加上

1

2 sin^{2} (θ) = 1 - cos (2 θ)

两边除以

2

sin^{2} (θ) = \frac{1 - cos ( 2 θ )}{2}

使用三角恒等式改写:

cos^{2} (θ) = \frac{1 + cos ( 2 θ )}{2}

使用倍角公式

cos (2 θ) = 2 cos^{2} (θ) - 1

交换两边

2 cos^{2} (θ) - 1 = cos (2 θ)

两边加上

1

2 sin^{2} (θ) = 1 + cos (2 θ)

两边除以

2

cos^{2} (θ) = \frac{1 + cos ( 2 θ )}{2}

tan^{2} (θ) = \frac{\frac{1 - c o s ( 2 θ )}{2}}{\frac{1 + c o s ( 2 θ )}{2}}

化简

tan^{2} (θ) = \frac{1 - cos ( 2 θ )}{1 + cos ( 2 θ )}

用

\frac{θ}{2}

替代

θ

tan^{2} (\frac{θ}{2}) = \frac{1 - cos ( 2 \cdot \frac{θ}{2} )}{1 + cos ( 2 \cdot \frac{θ}{2} )}

化简

tan^{2} (\frac{θ}{2}) = \frac{1 - cos ( θ )}{1 + cos ( θ )}

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

:

range [0, \frac{π}{2}] [\frac{π}{2}, π] quadrant I II tan positive negative

tan (\frac{θ}{2}) = \frac{1 - cos ( θ )}{1 + cos ( θ )}

= \frac{1 - cos ( \frac{π}{10} )}{1 + cos ( \frac{π}{10} )}

= \frac{1 - cos ( \frac{π}{10} )}{1 + cos ( \frac{π}{10} )}

使用三角恒等式改写:

cos (\frac{π}{10}) = \frac{2 5 + 5}{4}

cos (\frac{π}{10})

使用三角恒等式改写:

\frac{1 + cos ( \frac{π}{5} )}{2}

cos (\frac{π}{10})

将

cos (\frac{π}{10})

写为

cos (\frac{\frac{π}{5}}{2})

= cos (\frac{\frac{π}{5}}{2})

使用半角公式:

cos (\frac{θ}{2}) = \frac{1 + cos ( θ )}{2}

使用倍角公式

cos (2 θ) = 2 cos^{2} (θ) - 1

用

\frac{θ}{2}

替代

θ

cos (θ) = 2 cos^{2} (\frac{θ}{2}) - 1

交换两边

2 cos^{2} (\frac{θ}{2}) = 1 + cos (θ)

两边除以

2

cos^{2} (\frac{θ}{2}) = \frac{( 1 + cos ( θ ) )}{2}

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

:

range [0, \frac{π}{2}] [\frac{π}{2}, π] [π, \frac{3 π}{2}] [\frac{3 π}{2}, 2 π] quadrant I II III I V sin positive positive negative negative cos positive negative negative positive

cos (\frac{θ}{2}) = \frac{( 1 + cos ( θ ) )}{2}

= \frac{1 + cos ( \frac{π}{5} )}{2}

= \frac{1 + cos ( \frac{π}{5} )}{2}

使用三角恒等式改写:

cos (\frac{π}{5}) = \frac{5 + 1}{4}

cos (\frac{π}{5})

显示:

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \frac{1}{2}

使用以下积化和差公式:

2 sin (x) cos (y) = sin (x + y) - sin (x - y)

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

显示:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

利用以下特性:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

两边除以

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

代入

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

\frac{1}{2} = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

sin (\frac{3 π}{10}) = cos (\frac{π}{2} - \frac{3 π}{10})

\frac{1}{2} = cos (\frac{π}{2} - \frac{3 π}{10}) - sin (\frac{π}{10})

\frac{1}{2} = cos (\frac{π}{5}) - sin (\frac{π}{10})

显示:

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

使用因式分解法则:

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

a = cos (\frac{π}{5}) + sin (\frac{π}{10})

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = ((cos (\frac{π}{5}) + sin (\frac{π}{10})) + (cos (\frac{π}{5}) - sin (\frac{π}{10}))) ((cos (\frac{π}{5}) + sin (\frac{π}{10})) - (cos (\frac{π}{5}) - sin (\frac{π}{10})))

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 2 (2 cos (\frac{π}{5}) sin (\frac{π}{10}))

显示:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

利用以下特性:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

两边除以

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

代入

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 1

代入

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (\frac{1}{2})^{2} = 1

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} = 1

两边加上

\frac{1}{4}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} = \frac{5}{4}

在两侧开平方

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \pm \frac{5}{4}

cos (\frac{π}{5})

不能为负

sin (\frac{π}{10})

不能为负

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

以下方程式相加

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{2}

((cos (\frac{π}{5}) + sin (\frac{π}{10})) + (cos (\frac{π}{5}) - sin (\frac{π}{10}))) = (\frac{5}{2} + \frac{1}{2})

整理后得

cos (\frac{π}{5}) = \frac{5 + 1}{4}

= \frac{5 + 1}{4}

= \frac{1 + \frac{5 + 1}{4}}{2}

化简

\frac{1 + \frac{5 + 1}{4}}{2} : \frac{2 5 + 5}{4}

\frac{1 + \frac{5 + 1}{4}}{2}

\frac{1 + \frac{5 + 1}{4}}{2} = \frac{5 + 5}{8}

\frac{1 + \frac{5 + 1}{4}}{2}

化简

1 + \frac{5 + 1}{4} : \frac{5 + 5}{4}

1 + \frac{5 + 1}{4}

将项转换为分式:

1 = \frac{1 \cdot 4}{4}

= \frac{1 \cdot 4}{4} + \frac{5 + 1}{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{1 \cdot 4 + 5 + 1}{4}

1 \cdot 4 + 5 + 1 = 5 + 5

1 \cdot 4 + 5 + 1

数字相乘:

1 \cdot 4 = 4

= 4 + 5 + 1

数字相加:

4 + 1 = 5

= 5 + 5

= \frac{5 + 5}{4}

= \frac{\frac{5 + 5}{4}}{2}

使用分式法则:

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

= \frac{5 + 5}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{5 + 5}{8}

= \frac{5 + 5}{8}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{5 + 5}{8}

8 = 22

8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

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

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

= \frac{5 + 5}{2 2}

\frac{5 + 5}{2 2}

有理化:

\frac{2 5 + 5}{4}

\frac{5 + 5}{2 2}

乘以共轭根式

\frac{2}{2}

= \frac{5 + 5 2}{2 2 2}

222 = 4

222

使用指数法则:

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

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

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

同类项相加:

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

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

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

2 \cdot \frac{1}{2}

分式相乘:

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

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

约分:

2

= 1

= 2^{1 + 1}

数字相加:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2 5 + 5}{4}

= \frac{2 5 + 5}{4}

= \frac{2 5 + 5}{4}

= \frac{1 - \frac{2 5 + 5}{4}}{1 + \frac{2 5 + 5}{4}}

化简

\frac{1 - \frac{2 5 + 5}{4}}{1 + \frac{2 5 + 5}{4}} : - 32 5 + 5 - 10 5 + 5 + 11 + 45

\frac{1 - \frac{2 5 + 5}{4}}{1 + \frac{2 5 + 5}{4}}

\frac{1 - \frac{2 5 + 5}{4}}{1 + \frac{2 5 + 5}{4}} = \frac{4 - 2 5 + 5}{4 + 2 5 + 5}

\frac{1 - \frac{2 5 + 5}{4}}{1 + \frac{2 5 + 5}{4}}

化简

1 + \frac{2 5 + 5}{4} : \frac{4 + 2 5 + 5}{4}

1 + \frac{2 5 + 5}{4}

将项转换为分式:

1 = \frac{1 \cdot 4}{4}

= \frac{1 \cdot 4}{4} + \frac{2 5 + 5}{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{1 \cdot 4 + 2 5 + 5}{4}

数字相乘:

1 \cdot 4 = 4

= \frac{4 + 2 5 + 5}{4}

= \frac{1 - \frac{2 5 + 5}{4}}{\frac{4 + 2 5 + 5}{4}}

化简

1 - \frac{2 5 + 5}{4} : \frac{4 - 2 5 + 5}{4}

1 - \frac{2 5 + 5}{4}

将项转换为分式:

1 = \frac{1 \cdot 4}{4}

= \frac{1 \cdot 4}{4} - \frac{2 5 + 5}{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{1 \cdot 4 - 2 5 + 5}{4}

数字相乘:

1 \cdot 4 = 4

= \frac{4 - 2 5 + 5}{4}

= \frac{\frac{4 - 2 5 + 5}{4}}{\frac{4 + 2 5 + 5}{4}}

分式相除:

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

= \frac{( 4 - 2 5 + 5 ) \cdot 4}{4 ( 4 + 2 5 + 5 )}

约分:

4

= \frac{4 - 2 5 + 5}{4 + 2 5 + 5}

= \frac{4 - 2 5 + 5}{4 + 2 5 + 5}

\frac{4 - 2 5 + 5}{4 + 2 5 + 5} = - 32 5 + 5 - 10 5 + 5 + 11 + 45

\frac{4 - 2 5 + 5}{4 + 2 5 + 5}

乘以共轭根式

\frac{4 - 2 5 + 5}{4 - 2 5 + 5}

= \frac{( 4 - 2 5 + 5 ) ( 4 - 2 5 + 5 )}{( 4 + 2 5 + 5 ) ( 4 - 2 5 + 5 )}

(4 - 2 5 + 5) (4 - 2 5 + 5) = - 82 5 + 5 + 26 + 25

(4 - 2 5 + 5) (4 - 2 5 + 5)

使用指数法则:

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

(4 - 2 5 + 5) (4 - 2 5 + 5) = (4 - 2 5 + 5)^{1 + 1}

= (4 - 2 5 + 5)^{1 + 1}

数字相加:

1 + 1 = 2

= (4 - 2 5 + 5)^{2}

使用完全平方公式:

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

a = 4, b = 2 5 + 5

= 4^{2} - 2 \cdot 42 5 + 5 + (2 5 + 5)^{2}

化简

4^{2} - 2 \cdot 42 5 + 5 + (2 5 + 5)^{2} : - 82 5 + 5 + 26 + 25

4^{2} - 2 \cdot 42 5 + 5 + (2 5 + 5)^{2}

4^{2} = 16

4^{2}

4^{2} = 16

= 16

2 \cdot 42 5 + 5 = 82 5 + 5

2 \cdot 42 5 + 5

数字相乘:

2 \cdot 4 = 8

= 82 5 + 5

(2 5 + 5)^{2} = 2 (5 + 5)

(2 5 + 5)^{2}

使用指数法则:

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

= (2)^{2} (5 + 5)^{2}

(2)^{2} : 2

使用根式运算法则:

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

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

使用指数法则:

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

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

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

= 2

= 2 (5 + 5)^{2}

(5 + 5)^{2} : 5 + 5

使用根式运算法则:

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

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

使用指数法则:

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

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

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

= 5 + 5

= 2 (5 + 5)

= 16 - 82 5 + 5 + 2 (5 + 5)

乘开

2 (5 + 5) : 10 + 25

2 (5 + 5)

使用分配律:

a (b + c) = ab + a c

a = 2, b = 5, c = 5

= 2 \cdot 5 + 25

数字相乘:

2 \cdot 5 = 10

= 10 + 25

= 16 - 82 5 + 5 + 10 + 25

数字相加:

16 + 10 = 26

= - 82 5 + 5 + 26 + 25

= - 82 5 + 5 + 26 + 25

(4 + 2 5 + 5) (4 - 2 5 + 5) = 6 - 25

(4 + 2 5 + 5) (4 - 2 5 + 5)

2 5 + 5 = 10 + 25

2 5 + 5

使用根式运算法则:

a b = a \cdot b

2 5 + 5 = 2 (5 + 5)

= 2 (5 + 5)

乘开

2 (5 + 5) : 10 + 25

2 (5 + 5)

使用分配律:

a (b + c) = ab + a c

a = 2, b = 5, c = 5

= 2 \cdot 5 + 25

数字相乘:

2 \cdot 5 = 10

= 10 + 25

= 10 + 25

= (10 + 25 + 4) (- 2 5 + 5 + 4)

2 5 + 5 = 10 + 25

2 5 + 5

使用根式运算法则:

a b = a \cdot b

2 5 + 5 = 2 (5 + 5)

= 2 (5 + 5)

乘开

2 (5 + 5) : 10 + 25

2 (5 + 5)

使用分配律:

a (b + c) = ab + a c

a = 2, b = 5, c = 5

= 2 \cdot 5 + 25

数字相乘:

2 \cdot 5 = 10

= 10 + 25

= 10 + 25

= (10 + 25 + 4) (- 10 + 25 + 4)

使用平方差公式:

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

a = 4, b = 10 + 25

= 4^{2} - (10 + 25)^{2}

化简

4^{2} - (10 + 25)^{2} : 6 - 25

4^{2} - (10 + 25)^{2}

4^{2} = 16

4^{2}

4^{2} = 16

= 16

(10 + 25)^{2} = 10 + 25

(10 + 25)^{2}

使用根式运算法则:

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

= ((10 + 25)^{\frac{1}{2}})^{2}

使用指数法则:

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

= (10 + 25)^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

= 10 + 25

= 16 - (10 + 25)

- (10 + 25) : - 10 - 25

- (10 + 25)

打开括号

= - (10) - (25)

使用加减运算法则

+ (- a) = - a

= - 10 - 25

= 16 - 10 - 25

数字相减:

16 - 10 = 6

= 6 - 25

= 6 - 25

= \frac{- 8 2 5 + 5 + 26 + 2 5}{6 - 2 5}

分解

- 82 5 + 5 + 26 + 25 : 2 (- 42 5 + 5 + 13 + 5)

- 82 5 + 5 + 26 + 25

改写为

= - 2 \cdot 42 5 + 5 + 2 \cdot 13 + 25

因式分解出通项

2

= 2 (- 42 5 + 5 + 13 + 5)

= \frac{2 ( - 4 2 5 + 5 + 13 + 5 )}{6 - 2 5}

分解

6 - 25 : 2 (3 - 5)

6 - 25

改写为

= 2 \cdot 3 - 25

因式分解出通项

2

= 2 (3 - 5)

= \frac{2 ( - 4 2 5 + 5 + 13 + 5 )}{2 ( 3 - 5 )}

数字相除:

\frac{2}{2} = 1

= \frac{- 4 2 5 + 5 + 13 + 5}{( 3 - 5 )}

去除括号:

(a) = a

= \frac{- 4 2 5 + 5 + 13 + 5}{3 - 5}

乘以共轭根式

\frac{3 + 5}{3 + 5}

= \frac{( - 4 2 5 + 5 + 13 + 5 ) ( 3 + 5 )}{( 3 - 5 ) ( 3 + 5 )}

(- 42 5 + 5 + 13 + 5) (3 + 5) = - 122 5 + 5 - 410 5 + 5 + 44 + 165

(- 42 5 + 5 + 13 + 5) (3 + 5)

打开括号

= (- 42 5 + 5) \cdot 3 + (- 42 5 + 5) 5 + 13 \cdot 3 + 135 + 5 \cdot 3 + 55

使用加减运算法则

+ (- a) = - a

= - 4 \cdot 32 5 + 5 - 425 5 + 5 + 13 \cdot 3 + 135 + 35 + 55

化简

- 4 \cdot 32 5 + 5 - 425 5 + 5 + 13 \cdot 3 + 135 + 35 + 55 : - 122 5 + 5 - 410 5 + 5 + 44 + 165

- 4 \cdot 32 5 + 5 - 425 5 + 5 + 13 \cdot 3 + 135 + 35 + 55

同类项相加:

135 + 35 = 165

= - 4 \cdot 32 5 + 5 - 425 5 + 5 + 13 \cdot 3 + 165 + 55

4 \cdot 32 5 + 5 = 122 5 + 5

4 \cdot 32 5 + 5

数字相乘:

4 \cdot 3 = 12

= 122 5 + 5

425 5 + 5 = 410 5 + 5

425 5 + 5

使用根式运算法则:

a b = a \cdot b

25 5 + 5 = 2 \cdot 5 (5 + 5)

= 4 2 \cdot 5 (5 + 5)

数字相乘:

2 \cdot 5 = 10

= 4 10 (5 + 5)

使用根式运算法则:

n ab = n a n b,

假定

a \geq 0, b \geq 0

10 (5 + 5) = 10 5 + 5

= 410 5 + 5

13 \cdot 3 = 39

13 \cdot 3

数字相乘:

13 \cdot 3 = 39

= 39

55 = 5

55

使用根式运算法则:

a a = a

55 = 5

= 5

= - 122 5 + 5 - 410 5 + 5 + 39 + 165 + 5

数字相加:

39 + 5 = 44

= - 122 5 + 5 - 410 5 + 5 + 44 + 165

= - 122 5 + 5 - 410 5 + 5 + 44 + 165

(3 - 5) (3 + 5) = 4

(3 - 5) (3 + 5)

使用平方差公式:

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

a = 3, b = 5

= 3^{2} - (5)^{2}

化简

3^{2} - (5)^{2} : 4

3^{2} - (5)^{2}

3^{2} = 9

3^{2}

3^{2} = 9

= 9

(5)^{2} = 5

(5)^{2}

使用根式运算法则:

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

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

使用指数法则:

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

= 5^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

= 5

= 9 - 5

数字相减:

9 - 5 = 4

= 4

= 4

= \frac{- 12 2 5 + 5 - 4 10 5 + 5 + 44 + 16 5}{4}

分解

- 122 5 + 5 - 410 5 + 5 + 44 + 165 : 4 (- 32 5 + 5 - 10 5 + 5 + 11 + 45)

- 122 5 + 5 - 410 5 + 5 + 44 + 165

改写为

= - 4 \cdot 32 5 + 5 - 410 5 + 5 + 4 \cdot 11 + 4 \cdot 45

因式分解出通项

4

= 4 (- 32 5 + 5 - 10 5 + 5 + 11 + 45)

= \frac{4 ( - 3 2 5 + 5 - 10 5 + 5 + 11 + 4 5 )}{4}

数字相除:

\frac{4}{4} = 1

= - 32 5 + 5 - 10 5 + 5 + 11 + 45

= - 32 5 + 5 - 10 5 + 5 + 11 + 45

= - 32 5 + 5 - 10 5 + 5 + 11 + 45

= \frac{2 5 π - 3 2 5 + 5 - 10 5 + 5 + 11 + 4 5}{π}

化简

\frac{2 5 π - 3 2 5 + 5 - 10 5 + 5 + 11 + 4 5}{π} : \frac{2 5 π 4 - 3 2 5 + 5 - 10 5 + 5 + 11 + 4 5}{π}

\frac{2 5 π - 3 2 5 + 5 - 10 5 + 5 + 11 + 4 5}{π}

- 32 5 + 5 - 10 5 + 5 + 11 + 45 = 4 - 32 5 + 5 - 10 5 + 5 + 11 + 45

- 32 5 + 5 - 10 5 + 5 + 11 + 45

使用根式运算法则:

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

= ((- 32 5 + 5 - 10 5 + 5 + 11 + 45)^{\frac{1}{2}})^{\frac{1}{2}}

使用指数法则:

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

= (- 32 5 + 5 - 10 5 + 5 + 11 + 45)^{\frac{1}{2} \cdot \frac{1}{2}}

\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}

\frac{1}{2} \cdot \frac{1}{2}

分式相乘:

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

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

数字相乘:

1 \cdot 1 = 1

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

数字相乘:

2 \cdot 2 = 4

= \frac{1}{4}

= (- 32 5 + 5 - 10 5 + 5 + 11 + 45)^{\frac{1}{4}}

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

= 4 - 32 5 + 5 - 10 5 + 5 + 11 + 45

= \frac{2 5 π 4 - 3 2 5 + 5 - 10 5 + 5 + 11 + 4 5}{π}

= \frac{2 5 π 4 - 3 2 5 + 5 - 10 5 + 5 + 11 + 4 5}{π}

流行的例子

- 2 cos (12 0^{\circ})

cos (0.76)

cos (0.72)

cos (0.68)

cos(150)-sin(240)+tan(300)

cos (15 0^{\circ}) - sin (24 0^{\circ}) + tan (30 0^{\circ})