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受欢迎的三角函数 >

tan(2sqrt(x)-3)=-1

解答

tan (2 x - 3) = - 1

解答

x = \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4}

求解步骤

tan (2 x - 3) = - 1

tan (2 x - 3) = - 1

的通解

tan (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

2 x - 3 = \frac{3 π}{4} + πn

2 x - 3 = \frac{3 π}{4} + πn

解

2 x - 3 = \frac{3 π}{4} + πn : x = \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4} {n > - \frac{96 π + 24 π ^{2}}{32 π ^{2}}}

2 x - 3 = \frac{3 π}{4} + πn

在两边乘以

4

2 x \cdot 4 - 3 \cdot 4 = \frac{3 π}{4} \cdot 4 + πn \cdot 4

化简

8 x - 12 = 3 π + 4 πn

将

12

到右边

8 x - 12 = 3 π + 4 πn

两边加上

12

8 x - 12 + 12 = 3 π + 4 πn + 12

化简

8 x = 3 π + 4 πn + 12

8 x = 3 π + 4 πn + 12

两边除以

8

8 x = 3 π + 4 πn + 12

两边除以

8

\frac{8 x}{8} = \frac{3 π}{8} + \frac{4 πn}{8} + \frac{12}{8}

化简

\frac{8 x}{8} = \frac{3 π}{8} + \frac{4 πn}{8} + \frac{12}{8}

化简

\frac{8 x}{8} : x

\frac{8 x}{8}

数字相除:

\frac{8}{8} = 1

= x

化简

\frac{3 π}{8} + \frac{4 πn}{8} + \frac{12}{8} : \frac{3 π}{8} + \frac{3}{2} + \frac{πn}{2}

\frac{3 π}{8} + \frac{4 πn}{8} + \frac{12}{8}

对同类项分组

= \frac{3 π}{8} + \frac{12}{8} + \frac{4 πn}{8}

消掉

\frac{12}{8} : \frac{3}{2}

\frac{12}{8}

约分:

4

= \frac{3}{2}

= \frac{3 π}{8} + \frac{3}{2} + \frac{4 πn}{8}

消掉

\frac{4 πn}{8} : \frac{πn}{2}

\frac{4 πn}{8}

约分:

4

= \frac{πn}{2}

= \frac{3 π}{8} + \frac{3}{2} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{3}{2} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{3}{2} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{3}{2} + \frac{πn}{2}

两边进行平方:

x = \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4}

x = \frac{3 π}{8} + \frac{3}{2} + \frac{πn}{2}

(x)^{2} = (\frac{3 π}{8} + \frac{3}{2} + \frac{πn}{2})^{2}

展开

(x)^{2} : x

(x)^{2}

使用根式运算法则:

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

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

使用指数法则:

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

= x^{\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

= x

展开

(\frac{3 π}{8} + \frac{3}{2} + \frac{πn}{2})^{2} : \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4}

(\frac{3 π}{8} + \frac{3}{2} + \frac{πn}{2})^{2}

合并分式

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

使用法则

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

= \frac{3 + πn}{2}

= (\frac{3 π}{8} + \frac{πn + 3}{2})^{2}

使用完全平方公式:

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

a = \frac{3 π}{8}, b = \frac{3 + πn}{2}

= (\frac{3 π}{8})^{2} + 2 \cdot \frac{3 π}{8} \cdot \frac{3 + πn}{2} + (\frac{3 + πn}{2})^{2}

化简

(\frac{3 π}{8})^{2} + 2 \cdot \frac{3 π}{8} \cdot \frac{3 + πn}{2} + (\frac{3 + πn}{2})^{2} : \frac{9 π ^{2}}{64} + \frac{9 π + 3 π ^{2} n}{8} + \frac{9 + 6 πn + π ^{2} n ^{2}}{4}

(\frac{3 π}{8})^{2} + 2 \cdot \frac{3 π}{8} \cdot \frac{3 + πn}{2} + (\frac{3 + πn}{2})^{2}

(\frac{3 π}{8})^{2} = \frac{9 π ^{2}}{64}

(\frac{3 π}{8})^{2}

使用指数法则:

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

= \frac{( 3 π ) ^{2}}{8 ^{2}}

使用指数法则:

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

(3 π)^{2} = 3^{2} π^{2}

= \frac{3 ^{2} π ^{2}}{8 ^{2}}

整理后得

= \frac{9 π ^{2}}{64}

2 \cdot \frac{3 π}{8} \cdot \frac{3 + πn}{2} = \frac{9 π + 3 π ^{2} n}{8}

2 \cdot \frac{3 π}{8} \cdot \frac{3 + πn}{2}

分式相乘:

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

= \frac{3 π ( 3 + πn ) \cdot 2}{8 \cdot 2}

约分:

2

= \frac{3 π ( 3 + πn )}{8}

乘开

3 π (3 + πn) : 9 π + 3 π^{2} n

3 π (3 + πn)

使用分配律:

a (b + c) = ab + a c

a = 3 π, b = 3, c = πn

= 3 π 3 + 3 ππn

= 3 \cdot 3 π + 3 ππn

化简

3 \cdot 3 π + 3 ππn : 9 π + 3 π^{2} n

3 \cdot 3 π + 3 ππn

3 \cdot 3 π = 9 π

3 \cdot 3 π

数字相乘:

3 \cdot 3 = 9

= 9 π

3 ππn = 3 π^{2} n

3 ππn

使用指数法则:

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

ππ = π^{1 + 1}

= 3 π^{1 + 1} n

数字相加:

1 + 1 = 2

= 3 π^{2} n

= 9 π + 3 π^{2} n

= 9 π + 3 π^{2} n

= \frac{9 π + 3 π ^{2} n}{8}

(\frac{3 + πn}{2})^{2} = \frac{9 + 6 πn + π ^{2} n ^{2}}{4}

(\frac{3 + πn}{2})^{2}

使用指数法则:

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

= \frac{( 3 + πn ) ^{2}}{2 ^{2}}

(3 + πn)^{2} = 9 + 6 πn + π^{2} n^{2}

(3 + πn)^{2}

使用完全平方公式:

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

a = 3, b = πn

= 3^{2} + 2 \cdot 3 πn + (πn)^{2}

化简

3^{2} + 2 \cdot 3 πn + (πn)^{2} : 9 + 6 πn + π^{2} n^{2}

3^{2} + 2 \cdot 3 πn + (πn)^{2}

3^{2} = 9

= 9 + 2 \cdot 3 πn + (πn)^{2}

数字相乘:

2 \cdot 3 = 6

= 9 + 6 πn + (πn)^{2}

使用指数法则:

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

= 9 + 6 πn + π^{2} n^{2}

= 9 + 6 πn + π^{2} n^{2}

= \frac{9 + 6 πn + π ^{2} n ^{2}}{2 ^{2}}

2^{2} = 4

= \frac{9 + 6 πn + π ^{2} n ^{2}}{4}

= \frac{9 π ^{2}}{64} + \frac{3 π ^{2} n + 9 π}{8} + \frac{π ^{2} n ^{2} + 6 πn + 9}{4}

= \frac{9 π ^{2}}{64} + \frac{9 π + 3 π ^{2} n}{8} + \frac{9 + 6 πn + π ^{2} n ^{2}}{4}

使用分式法则:

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

\frac{9 π + 3 π ^{2} n}{8} = \frac{9 π}{8} + \frac{3 π ^{2} n}{8}

= \frac{9 π ^{2}}{64} + \frac{9 π}{8} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2} + 6 πn + 9}{4}

使用分式法则:

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

\frac{9 + 6 πn + π ^{2} n ^{2}}{4} = \frac{9}{4} + \frac{6 πn}{4} + \frac{π ^{2} n ^{2}}{4}

= \frac{9 π ^{2}}{64} + \frac{9 π}{8} + \frac{3 π ^{2} n}{8} + \frac{9}{4} + \frac{6 πn}{4} + \frac{π ^{2} n ^{2}}{4}

对同类项分组

= \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{6 πn}{4} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4}

消掉

\frac{6 πn}{4} : \frac{3 πn}{2}

\frac{6 πn}{4}

约分:

2

= \frac{3 πn}{2}

= \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4}

x = \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4}

x = \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4}

验证解:

x = \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4} {n > - \frac{96 π + 24 π ^{2}}{32 π ^{2}}}

将它们代入

2 x - 3 = \frac{3 π}{4} + πn

检验解是否符合

去除与方程不符的解。

代入

x = \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4} : 2 \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4} - 3 = \frac{3 π}{4} + πn \Rightarrow n > - \frac{96 π + 24 π ^{2}}{32 π ^{2}}

2 \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4} - 3 = \frac{3 π}{4} + πn

在两边乘以

4

2 \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4} \cdot 4 - 3 \cdot 4 = \frac{3 π}{4} \cdot 4 + πn \cdot 4

化简

8 \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4} - 12 = 3 π + 4 πn

去除平方根

8 \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4} - 12 = 3 π + 4 πn

两边加上

12

8 \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4} - 12 + 12 = 3 π + 4 πn + 12

化简

8 \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4} = 3 π + 4 πn + 12

两边进行平方:

16 π^{2} n^{2} + 144 + 24 π^{2} n + 72 π + 9 π^{2} + 96 πn = 9 π^{2} + 24 π^{2} n + 72 π + 16 π^{2} n^{2} + 96 πn + 144

8 \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4} - 12 = 3 π + 4 πn

(8 \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4})^{2} = (3 π + 4 πn + 12)^{2}

展开

(8 \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4})^{2} : 16 π^{2} n^{2} + 144 + 24 π^{2} n + 72 π + 9 π^{2} + 96 πn

(8 \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4})^{2}

使用指数法则:

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

= 8^{2} (\frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4})^{2}

(\frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4})^{2} : \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4}

使用根式运算法则:

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

= ((\frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4})^{\frac{1}{2}})^{2}

使用指数法则:

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

= (\frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4})^{\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

= \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4}

= 8^{2} (\frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4})

8^{2} = 64

= 64 (\frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4})

展开

64 (\frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4}) : 16 π^{2} n^{2} + 144 + 24 π^{2} n + 72 π + 9 π^{2} + 96 πn

64 (\frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4})

化简

\frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4} : \frac{π ^{2} n ^{2} + 9}{4} + \frac{3 π ^{2} n + 9 π}{8} + \frac{3 πn}{2} + \frac{9 π ^{2}}{64}

\frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4}

合并分式

\frac{9}{4} + \frac{π ^{2} n ^{2}}{4} : \frac{9 + π ^{2} n ^{2}}{4}

使用法则

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

= \frac{9 + π ^{2} n ^{2}}{4}

= \frac{π ^{2} n ^{2} + 9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8}

合并分式

\frac{9 π}{8} + \frac{3 π ^{2} n}{8} : \frac{9 π + 3 π ^{2} n}{8}

使用法则

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

= \frac{9 π + 3 π ^{2} n}{8}

= \frac{π ^{2} n ^{2} + 9}{4} + \frac{3 π ^{2} n + 9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2}

= 64 (\frac{π ^{2} n ^{2} + 9}{4} + \frac{3 π ^{2} n + 9 π}{8} + \frac{3 πn}{2} + \frac{9 π ^{2}}{64})

打开括号

= 64 \cdot \frac{9 + π ^{2} n ^{2}}{4} + 64 \cdot \frac{9 π + 3 π ^{2} n}{8} + 64 \cdot \frac{9 π ^{2}}{64} + 64 \cdot \frac{3 πn}{2}

化简

64 \cdot \frac{9 + π ^{2} n ^{2}}{4} + 64 \cdot \frac{9 π + 3 π ^{2} n}{8} + 64 \cdot \frac{9 π ^{2}}{64} + 64 \cdot \frac{3 πn}{2} : 16 π^{2} n^{2} + 144 + 24 π^{2} n + 72 π + 9 π^{2} + 96 πn

64 \cdot \frac{9 + π ^{2} n ^{2}}{4} + 64 \cdot \frac{9 π + 3 π ^{2} n}{8} + 64 \cdot \frac{9 π ^{2}}{64} + 64 \cdot \frac{3 πn}{2}

64 \cdot \frac{9 + π ^{2} n ^{2}}{4} = 16 π^{2} n^{2} + 144

64 \cdot \frac{9 + π ^{2} n ^{2}}{4}

分式相乘:

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

= \frac{( 9 + π ^{2} n ^{2} ) \cdot 64}{4}

数字相除:

\frac{64}{4} = 16

= 16 (π^{2} n^{2} + 9)

使用分配律:

a (b + c) = ab + a c

a = 16, b = π^{2} n^{2}, c = 9

= 16 π^{2} n^{2} + 16 \cdot 9

数字相乘:

16 \cdot 9 = 144

= 16 π^{2} n^{2} + 144

64 \cdot \frac{9 π + 3 π ^{2} n}{8} = 24 π^{2} n + 72 π

64 \cdot \frac{9 π + 3 π ^{2} n}{8}

分式相乘:

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

= \frac{( 9 π + 3 π ^{2} n ) \cdot 64}{8}

数字相除:

\frac{64}{8} = 8

= 8 (3 π^{2} n + 9 π)

使用分配律:

a (b + c) = ab + a c

a = 8, b = 3 π^{2} n, c = 9 π

= 8 \cdot 3 π^{2} n + 8 \cdot 9 π

化简

8 \cdot 3 π^{2} n + 8 \cdot 9 π : 24 π^{2} n + 72 π

8 \cdot 3 π^{2} n + 8 \cdot 9 π

数字相乘:

8 \cdot 3 = 24

= 24 π^{2} n + 8 \cdot 9 π

数字相乘:

8 \cdot 9 = 72

= 24 π^{2} n + 72 π

= 24 π^{2} n + 72 π

64 \cdot \frac{9 π ^{2}}{64} = 9 π^{2}

64 \cdot \frac{9 π ^{2}}{64}

分式相乘:

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

= \frac{9 π ^{2} \cdot 64}{64}

约分:

64

= 9 π^{2}

64 \cdot \frac{3 πn}{2} = 96 πn

64 \cdot \frac{3 πn}{2}

分式相乘:

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

= \frac{3 πn \cdot 64}{2}

数字相乘:

3 \cdot 64 = 192

= \frac{192 πn}{2}

数字相除:

\frac{192}{2} = 96

= 96 πn

= 16 π^{2} n^{2} + 144 + 24 π^{2} n + 72 π + 9 π^{2} + 96 πn

= 16 π^{2} n^{2} + 144 + 24 π^{2} n + 72 π + 9 π^{2} + 96 πn

= 16 π^{2} n^{2} + 144 + 24 π^{2} n + 72 π + 9 π^{2} + 96 πn

展开

(3 π + 4 πn + 12)^{2} : 9 π^{2} + 24 π^{2} n + 72 π + 16 π^{2} n^{2} + 96 πn + 144

(3 π + 4 πn + 12)^{2}

(3 π + 4 πn + 12)^{2} = (3 π + 4 πn + 12) (3 π + 4 πn + 12)

= (3 π + 4 πn + 12) (3 π + 4 πn + 12)

乘开

(3 π + 4 πn + 12) (3 π + 4 πn + 12) : 9 π^{2} + 24 π^{2} n + 72 π + 16 π^{2} n^{2} + 96 πn + 144

(3 π + 4 πn + 12) (3 π + 4 πn + 12)

打开括号

= 3 π 3 π + 3 π 4 πn + 3 π 12 + 4 πn \cdot 3 π + 4 πn \cdot 4 πn + 4 πn \cdot 12 + 12 \cdot 3 π + 12 \cdot 4 πn + 12 \cdot 12

= 3 \cdot 3 ππ + 3 \cdot 4 ππn + 3 \cdot 12 π + 4 \cdot 3 ππn + 4 \cdot 4 ππnn + 4 \cdot 12 πn + 12 \cdot 3 π + 12 \cdot 4 πn + 12 \cdot 12

化简

3 \cdot 3 ππ + 3 \cdot 4 ππn + 3 \cdot 12 π + 4 \cdot 3 ππn + 4 \cdot 4 ππnn + 4 \cdot 12 πn + 12 \cdot 3 π + 12 \cdot 4 πn + 12 \cdot 12 : 9 π^{2} + 24 π^{2} n + 72 π + 16 π^{2} n^{2} + 96 πn + 144

3 \cdot 3 ππ + 3 \cdot 4 ππn + 3 \cdot 12 π + 4 \cdot 3 ππn + 4 \cdot 4 ππnn + 4 \cdot 12 πn + 12 \cdot 3 π + 12 \cdot 4 πn + 12 \cdot 12

同类项相加:

3 \cdot 12 π + 12 \cdot 3 π = 2 \cdot 12 \cdot 3 π

= 3 \cdot 3 ππ + 3 \cdot 4 ππn + 2 \cdot 12 \cdot 3 π + 4 \cdot 3 ππn + 4 \cdot 4 ππnn + 4 \cdot 12 πn + 12 \cdot 4 πn + 12 \cdot 12

同类项相加:

4 \cdot 12 πn + 12 \cdot 4 πn = 2 \cdot 12 \cdot 4 πn

= 3 \cdot 3 ππ + 3 \cdot 4 ππn + 2 \cdot 12 \cdot 3 π + 4 \cdot 3 ππn + 4 \cdot 4 ππnn + 2 \cdot 12 \cdot 4 πn + 12 \cdot 12

同类项相加:

3 \cdot 4 ππn + 4 \cdot 3 ππn = 2 \cdot 4 \cdot 3 ππn

= 3 \cdot 3 ππ + 2 \cdot 4 \cdot 3 ππn + 2 \cdot 12 \cdot 3 π + 4 \cdot 4 ππnn + 2 \cdot 12 \cdot 4 πn + 12 \cdot 12

3 \cdot 3 ππ = 9 π^{2}

3 \cdot 3 ππ

数字相乘:

3 \cdot 3 = 9

= 9 ππ

使用指数法则:

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

ππ = π^{1 + 1}

= 9 π^{1 + 1}

数字相加:

1 + 1 = 2

= 9 π^{2}

2 \cdot 4 \cdot 3 ππn = 24 π^{2} n

2 \cdot 4 \cdot 3 ππn

数字相乘:

2 \cdot 4 \cdot 3 = 24

= 24 ππn

使用指数法则:

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

ππ = π^{1 + 1}

= 24 π^{1 + 1} n

数字相加:

1 + 1 = 2

= 24 π^{2} n

2 \cdot 12 \cdot 3 π = 72 π

2 \cdot 12 \cdot 3 π

数字相乘:

2 \cdot 12 \cdot 3 = 72

= 72 π

4 \cdot 4 ππnn = 16 π^{2} n^{2}

4 \cdot 4 ππnn

数字相乘:

4 \cdot 4 = 16

= 16 ππnn

使用指数法则:

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

nn = n^{1 + 1}

= 16 ππ n^{1 + 1}

数字相加:

1 + 1 = 2

= 16 ππ n^{2}

使用指数法则:

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

ππ = π^{1 + 1}

= 16 π^{1 + 1} n^{2}

数字相加:

1 + 1 = 2

= 16 π^{2} n^{2}

2 \cdot 12 \cdot 4 πn = 96 πn

2 \cdot 12 \cdot 4 πn

数字相乘:

2 \cdot 12 \cdot 4 = 96

= 96 πn

12 \cdot 12 = 144

12 \cdot 12

数字相乘:

12 \cdot 12 = 144

= 144

= 9 π^{2} + 24 π^{2} n + 72 π + 16 π^{2} n^{2} + 96 πn + 144

= 9 π^{2} + 24 π^{2} n + 72 π + 16 π^{2} n^{2} + 96 πn + 144

= 9 π^{2} + 24 π^{2} n + 72 π + 16 π^{2} n^{2} + 96 πn + 144

16 π^{2} n^{2} + 144 + 24 π^{2} n + 72 π + 9 π^{2} + 96 πn = 9 π^{2} + 24 π^{2} n + 72 π + 16 π^{2} n^{2} + 96 πn + 144

16 π^{2} n^{2} + 144 + 24 π^{2} n + 72 π + 9 π^{2} + 96 πn = 9 π^{2} + 24 π^{2} n + 72 π + 16 π^{2} n^{2} + 96 πn + 144

16 π^{2} n^{2} + 144 + 24 π^{2} n + 72 π + 9 π^{2} + 96 πn = 9 π^{2} + 24 π^{2} n + 72 π + 16 π^{2} n^{2} + 96 πn + 144

解

16 π^{2} n^{2} + 144 + 24 π^{2} n + 72 π + 9 π^{2} + 96 πn = 9 π^{2} + 24 π^{2} n + 72 π + 16 π^{2} n^{2} + 96 πn + 144 :

对所有

n

为真

16 π^{2} n^{2} + 144 + 24 π^{2} n + 72 π + 9 π^{2} + 96 πn = 9 π^{2} + 24 π^{2} n + 72 π + 16 π^{2} n^{2} + 96 πn + 144

两边减去

16 π^{2} n^{2} + 144 + 24 π^{2} n + 72 π + 9 π^{2} + 96 πn

16 π^{2} n^{2} + 144 + 24 π^{2} n + 72 π + 9 π^{2} + 96 πn - (16 π^{2} n^{2} + 144 + 24 π^{2} n + 72 π + 9 π^{2} + 96 πn) = 9 π^{2} + 24 π^{2} n + 72 π + 16 π^{2} n^{2} + 96 πn + 144 - (16 π^{2} n^{2} + 144 + 24 π^{2} n + 72 π + 9 π^{2} + 96 πn)

化简

0 = 0

两侧相等

对所有 n 为真

对所有 n 为真

验证解:

n < - \frac{96 π + 24 π ^{2}}{32 π ^{2}}

假

, n = - \frac{96 π + 24 π ^{2}}{32 π ^{2}}

假

, n > - \frac{96 π + 24 π ^{2}}{32 π ^{2}}

真

2 \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4} - 3 = \frac{3 π}{4} + πn

将定义域区间与解区间合并:

对所有 n 为真

找到函数区间:

n < - \frac{96 π + 24 π ^{2}}{32 π ^{2}}, n = - \frac{96 π + 24 π ^{2}}{32 π ^{2}}, n > - \frac{96 π + 24 π ^{2}}{32 π ^{2}}

2 \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4} - 3 = \frac{3 π}{4} + πn

找到偶数根参量零值:

解

\frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4} = 0 : n = - \frac{96 π + 24 π ^{2}}{32 π ^{2}}

\frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4} = 0

找到

4, 8, 64, 2

的最小公倍数:

64

4, 8, 64, 2

最小公倍数 (LCM)

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

8

质因数分解:

2 \cdot 2 \cdot 2

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

64

质因数分解:

2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

64

64

除以

264 = 32 \cdot 2

= 2 \cdot 32

32

除以

232 = 16 \cdot 2

= 2 \cdot 2 \cdot 16

16

除以

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

计算出由至少在以下一个数字中出现的因数组成的数字:

4, 8, 64, 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

数字相乘:

2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 64

= 64

乘以最小公倍数=

64

\frac{9}{4} \cdot 64 + \frac{9 π}{8} \cdot 64 + \frac{9 π ^{2}}{64} \cdot 64 + \frac{3 πn}{2} \cdot 64 + \frac{3 π ^{2} n}{8} \cdot 64 + \frac{π ^{2} n ^{2}}{4} \cdot 64 = 0 \cdot 64

化简

16 π^{2} n^{2} + (96 π + 24 π^{2}) n + 144 + 72 π + 9 π^{2} = 0

使用求根公式求解

16 π^{2} n^{2} + (96 π + 24 π^{2}) n + 144 + 72 π + 9 π^{2} = 0

二次方程求根公式:

若

a = 16 π^{2}, b = 96 π + 24 π^{2}, c = 144 + 72 π + 9 π^{2}

n_{1, 2} = \frac{- ( 96 π + 24 π ^{2} ) \pm ( 96 π + 24 π ^{2} ) ^{2} - 4 \cdot 16 π ^{2} ( 144 + 72 π + 9 π ^{2} )}{2 \cdot 16 π ^{2}}

n_{1, 2} = \frac{- ( 96 π + 24 π ^{2} ) \pm ( 96 π + 24 π ^{2} ) ^{2} - 4 \cdot 16 π ^{2} ( 144 + 72 π + 9 π ^{2} )}{2 \cdot 16 π ^{2}}

(96 π + 24 π^{2})^{2} - 4 \cdot 16 π^{2} (144 + 72 π + 9 π^{2}) = 0

(96 π + 24 π^{2})^{2} - 4 \cdot 16 π^{2} (144 + 72 π + 9 π^{2})

数字相乘:

4 \cdot 16 = 64

= (24 π^{2} + 96 π)^{2} - 64 π^{2} (9 π^{2} + 72 π + 144)

(96 π + 24 π^{2})^{2} : 9216 π^{2} + 4608 π^{3} + 576 π^{4}

使用完全平方公式:

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

a = 96 π, b = 24 π^{2}

= (96 π)^{2} + 2 \cdot 96 π 24 π^{2} + (24 π^{2})^{2}

化简

(96 π)^{2} + 2 \cdot 96 π 24 π^{2} + (24 π^{2})^{2} : 9216 π^{2} + 4608 π^{3} + 576 π^{4}

(96 π)^{2} + 2 \cdot 96 π 24 π^{2} + (24 π^{2})^{2}

(96 π)^{2} = 9216 π^{2}

(96 π)^{2}

使用指数法则:

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

= 9 6^{2} π^{2}

9 6^{2} = 9216

= 9216 π^{2}

2 \cdot 96 π 24 π^{2} = 4608 π^{3}

2 \cdot 96 π 24 π^{2}

数字相乘:

2 \cdot 96 \cdot 24 = 4608

= 4608 π^{2} π

使用指数法则:

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

π π^{2} = π^{1 + 2}

= 4608 π^{1 + 2}

数字相加:

1 + 2 = 3

= 4608 π^{3}

(24 π^{2})^{2} = 576 π^{4}

(24 π^{2})^{2}

使用指数法则:

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

= 2 4^{2} (π^{2})^{2}

(π^{2})^{2} : π^{4}

使用指数法则:

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

= π^{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= π^{4}

= 2 4^{2} π^{4}

2 4^{2} = 576

= 576 π^{4}

= 9216 π^{2} + 4608 π^{3} + 576 π^{4}

= 9216 π^{2} + 4608 π^{3} + 576 π^{4}

= 9216 π^{2} + 4608 π^{3} + 576 π^{4} - 64 π^{2} (144 + 72 π + 9 π^{2})

乘开

- 64 π^{2} (144 + 72 π + 9 π^{2}) : - 9216 π^{2} - 4608 π^{3} - 576 π^{4}

- 64 π^{2} (144 + 72 π + 9 π^{2})

打开括号

= (- 64 π^{2}) \cdot 144 + (- 64 π^{2}) \cdot 72 π + (- 64 π^{2}) \cdot 9 π^{2}

使用加减运算法则

+ (- a) = - a

= - 64 \cdot 144 π^{2} - 64 \cdot 72 π^{2} π - 64 \cdot 9 π^{2} π^{2}

化简

- 64 \cdot 144 π^{2} - 64 \cdot 72 π^{2} π - 64 \cdot 9 π^{2} π^{2} : - 9216 π^{2} - 4608 π^{3} - 576 π^{4}

- 64 \cdot 144 π^{2} - 64 \cdot 72 π^{2} π - 64 \cdot 9 π^{2} π^{2}

64 \cdot 144 π^{2} = 9216 π^{2}

64 \cdot 144 π^{2}

数字相乘:

64 \cdot 144 = 9216

= 9216 π^{2}

64 \cdot 72 π^{2} π = 4608 π^{3}

64 \cdot 72 π^{2} π

数字相乘:

64 \cdot 72 = 4608

= 4608 π^{2} π

使用指数法则:

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

π^{2} π = π^{2 + 1}

= 4608 π^{2 + 1}

数字相加:

2 + 1 = 3

= 4608 π^{3}

64 \cdot 9 π^{2} π^{2} = 576 π^{4}

64 \cdot 9 π^{2} π^{2}

数字相乘:

64 \cdot 9 = 576

= 576 π^{2} π^{2}

使用指数法则:

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

π^{2} π^{2} = π^{2 + 2}

= 576 π^{2 + 2}

数字相加:

2 + 2 = 4

= 576 π^{4}

= - 9216 π^{2} - 4608 π^{3} - 576 π^{4}

= - 9216 π^{2} - 4608 π^{3} - 576 π^{4}

= 9216 π^{2} + 4608 π^{3} + 576 π^{4} - 9216 π^{2} - 4608 π^{3} - 576 π^{4}

化简

9216 π^{2} + 4608 π^{3} + 576 π^{4} - 9216 π^{2} - 4608 π^{3} - 576 π^{4} : 0

9216 π^{2} + 4608 π^{3} + 576 π^{4} - 9216 π^{2} - 4608 π^{3} - 576 π^{4}

对同类项分组

= 576 π^{4} - 576 π^{4} + 4608 π^{3} - 4608 π^{3} + 9216 π^{2} - 9216 π^{2}

同类项相加:

9216 π^{2} - 9216 π^{2} = 0

= 576 π^{4} - 576 π^{4} + 4608 π^{3} - 4608 π^{3}

同类项相加:

4608 π^{3} - 4608 π^{3} = 0

= 576 π^{4} - 576 π^{4}

同类项相加:

576 π^{4} - 576 π^{4} = 0

= 0

= 0

n_{1, 2} = \frac{- ( 96 π + 24 π ^{2} ) \pm 0}{2 \cdot 16 π ^{2}}

n = \frac{- ( 96 π + 24 π ^{2} )}{2 \cdot 16 π ^{2}}

\frac{- ( 96 π + 24 π ^{2} )}{2 \cdot 16 π ^{2}} = - \frac{96 π + 24 π ^{2}}{32 π ^{2}}

\frac{- ( 96 π + 24 π ^{2} )}{2 \cdot 16 π ^{2}}

数字相乘:

2 \cdot 16 = 32

= \frac{- ( 24 π ^{2} + 96 π )}{32 π ^{2}}

使用分式法则:

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

= - \frac{( 96 π + 24 π ^{2} )}{32 π ^{2}}

去除括号:

(a) = a

= - \frac{96 π + 24 π ^{2}}{32 π ^{2}}

n = - \frac{96 π + 24 π ^{2}}{32 π ^{2}}

二次方程组的解是:

n = - \frac{96 π + 24 π ^{2}}{32 π ^{2}}

n = - \frac{96 π + 24 π ^{2}}{32 π ^{2}}

区间围绕零点来定义:

n < - \frac{96 π + 24 π ^{2}}{32 π ^{2}}, n = - \frac{96 π + 24 π ^{2}}{32 π ^{2}}, n > - \frac{96 π + 24 π ^{2}}{32 π ^{2}}

将区间与定义域合并

n < - \frac{96 π + 24 π ^{2}}{32 π ^{2}}, n = - \frac{96 π + 24 π ^{2}}{32 π ^{2}}, n > - \frac{96 π + 24 π ^{2}}{32 π ^{2}}

将它们代入

2 \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4} - 3 = \frac{3 π}{4} + πn

检验解是否符合

去除与方程不符的解。

代入

n < - \frac{96 π + 24 π ^{2}}{32 π ^{2}} : 2 \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4} - 3 \neq = \frac{3 π}{4} + πn \Rightarrow

假

解是

n > - \frac{96 π + 24 π ^{2}}{32 π ^{2}}

解是

x = \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4} {n > - \frac{96 π + 24 π ^{2}}{32 π ^{2}}}

x = \frac{9}{4} + \frac{9 π}{8} + \frac{9 π ^{2}}{64} + \frac{3 πn}{2} + \frac{3 π ^{2} n}{8} + \frac{π ^{2} n ^{2}}{4}

作图

流行的例子

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sin(θ)=-27/190

sin (θ) = - \frac{27}{190}

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