展开 ((x-1)/x)^8

解答

展开

(\frac{x - 1}{x})^{8}

1 - \frac{8}{x} + \frac{28}{x ^{2}} - \frac{56}{x ^{3}} + \frac{70}{x ^{4}} - \frac{56}{x ^{5}} + \frac{28}{x ^{6}} - \frac{8}{x ^{7}} + \frac{1}{x ^{8}}

求解步骤

(\frac{x - 1}{x})^{8}

使用指数法则:

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

= \frac{( x - 1 ) ^{8}}{x ^{8}}

(x - 1)^{8} = x^{8} - 8 x^{7} + 28 x^{6} - 56 x^{5} + 70 x^{4} - 56 x^{3} + 28 x^{2} - 8 x + 1

= \frac{x ^{8} - 8 x ^{7} + 28 x ^{6} - 56 x ^{5} + 70 x ^{4} - 56 x ^{3} + 28 x ^{2} - 8 x + 1}{x ^{8}}

使用分式法则:

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

\frac{x ^{8} - 8 x ^{7} + 28 x ^{6} - 56 x ^{5} + 70 x ^{4} - 56 x ^{3} + 28 x ^{2} - 8 x + 1}{x ^{8}} = \frac{x ^{8}}{x ^{8}} - \frac{8 x ^{7}}{x ^{8}} + \frac{28 x ^{6}}{x ^{8}} - \frac{56 x ^{5}}{x ^{8}} + \frac{70 x ^{4}}{x ^{8}} - \frac{56 x ^{3}}{x ^{8}} + \frac{28 x ^{2}}{x ^{8}} - \frac{8 x}{x ^{8}} + \frac{1}{x ^{8}}

= \frac{x ^{8}}{x ^{8}} - \frac{8 x ^{7}}{x ^{8}} + \frac{28 x ^{6}}{x ^{8}} - \frac{56 x ^{5}}{x ^{8}} + \frac{70 x ^{4}}{x ^{8}} - \frac{56 x ^{3}}{x ^{8}} + \frac{28 x ^{2}}{x ^{8}} - \frac{8 x}{x ^{8}} + \frac{1}{x ^{8}}

使用法则

\frac{a}{a} = 1

\frac{x ^{8}}{x ^{8}} = 1

= 1 - \frac{8 x ^{7}}{x ^{8}} + \frac{28 x ^{6}}{x ^{8}} - \frac{56 x ^{5}}{x ^{8}} + \frac{70 x ^{4}}{x ^{8}} - \frac{56 x ^{3}}{x ^{8}} + \frac{28 x ^{2}}{x ^{8}} - \frac{8 x}{x ^{8}} + \frac{1}{x ^{8}}

\frac{8 x ^{7}}{x ^{8}} = \frac{8}{x}

\frac{28 x ^{6}}{x ^{8}} = \frac{28}{x ^{2}}

\frac{56 x ^{5}}{x ^{8}} = \frac{56}{x ^{3}}

\frac{70 x ^{4}}{x ^{8}} = \frac{70}{x ^{4}}

\frac{56 x ^{3}}{x ^{8}} = \frac{56}{x ^{5}}

\frac{28 x ^{2}}{x ^{8}} = \frac{28}{x ^{6}}

\frac{8 x}{x ^{8}} = \frac{8}{x ^{7}}

= 1 - \frac{8}{x} + \frac{28}{x ^{2}} - \frac{56}{x ^{3}} + \frac{70}{x ^{4}} - \frac{56}{x ^{5}} + \frac{28}{x ^{6}} - \frac{8}{x ^{7}} + \frac{1}{x ^{8}}

e x p an d 6 \cdot \frac{6}{8}

s im pl i f y \frac{164.6}{( s + 1 ) ( s + 2 )}

e x p an d \frac{3 ^{x} ( 3 ^{h} - 1 )}{h}

e x p an d (x (\frac{d y}{d x}) + y)^{2}

e x p an d \frac{( s + 3 )}{( s + 1 ) ^{2}}