展开 ((4x-1)^3)/(64)

解答

展开

\frac{( 4 x - 1 ) ^{3}}{64}

x^{3} - \frac{3 x ^{2}}{4} + \frac{3 x}{16} - \frac{1}{64}

求解步骤

\frac{( 4 x - 1 ) ^{3}}{64}

(4 x - 1)^{3} = 64 x^{3} - 48 x^{2} + 12 x - 1

= \frac{64 x ^{3} - 48 x ^{2} + 12 x - 1}{64}

使用分式法则:

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

\frac{64 x ^{3} - 48 x ^{2} + 12 x - 1}{64} = \frac{64 x ^{3}}{64} - \frac{48 x ^{2}}{64} + \frac{12 x}{64} - \frac{1}{64}

= \frac{64 x ^{3}}{64} - \frac{48 x ^{2}}{64} + \frac{12 x}{64} - \frac{1}{64}

数字相除:

\frac{64}{64} = 1

= x^{3} - \frac{48 x ^{2}}{64} + \frac{12 x}{64} - \frac{1}{64}

消掉

\frac{48 x ^{2}}{64} : \frac{3 x ^{2}}{4}

= x^{3} - \frac{3 x ^{2}}{4} + \frac{12 x}{64} - \frac{1}{64}

消掉

\frac{12 x}{64} : \frac{3 x}{16}

= x^{3} - \frac{3 x ^{2}}{4} + \frac{3 x}{16} - \frac{1}{64}

e x p an d \frac{[ ( ab - c d ) \times e f ]}{g h}

s im pl i f y ((2 x + 1) sin (x) - (x - 1) (x + 2) cos (x)) sin (x)

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

e x p an d e^{x} (1 - 3 e^{- 2 x})

e x p an d \frac{4 ( 1 - 9 x ^{2} )}{( 1 + 3 x ^{2} ) ^{3}}