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受欢迎的代数 >

化简 1/(w(w+i)^2)

解答

化简

\frac{1}{w ( w + i ) ^{2}}

解答

\frac{w ^{2} - 1}{w ( w ^{4} + 2 w ^{2} + 1 )} - \frac{2}{w ^{4} + 2 w ^{2} + 1} i

求解步骤

\frac{1}{w ( w + i ) ^{2}}

展开

(w + i)^{2} : w^{2} + 2 w i - 1

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

将

w^{2} + 2 w i - 1

改写成标准复数形式:

(w^{2} - 1) + 2 w i

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

乘以共轭根式

\frac{( w ^{2} - 1 ) - 2 w i}{( w ^{2} - 1 ) - 2 w i}

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

化简

\frac{1 \cdot ( ( w ^{2} - 1 ) - 2 w i )}{w ( ( w ^{2} - 1 ) + 2 w i ) ( ( w ^{2} - 1 ) - 2 w i )} : \frac{w ^{2} - 1 - 2 w i}{w ( w ^{4} + 2 w ^{2} + 1 )}

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

将

w^{2} - 1 - 2 w i

改写成标准复数形式:

(w^{2} - 1) - 2 w i

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

使用分式法则:

\frac{a + bi}{c} = \frac{a}{c} + \frac{b}{c} i

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

消掉

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

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

流行的例子

展开 (z+Δz)^2

e x p an d (z + Δ z)^{2}

展开 (\sqrt[3]{2(x+h)+6}-\sqrt[3]{2x+6})/h

e x p an d \frac{3 2 ( x + h ) + 6 - 3 2 x + 6}{h}

展开-(7(3+sqrt(3)))/6

e x p an d - \frac{7 ( 3 + 3 )}{6}

d/(dv)((p)(p+(2.4)/(v^2))(v-8.3))

\frac{d}{d v} ((p) (p + \frac{2.4}{v ^{2}}) (v - 8.3))

(d^3)/(dt^3)((s)((3+2t)^3))

\frac{d ^{3}}{d t ^{3}} ((s) ((3 + 2 t)^{3}))