展开 (x(x-1)(x-2)(x-3))/(24)

解答

展开

\frac{x ( x - 1 ) ( x - 2 ) ( x - 3 )}{24}

\frac{x ^{4}}{24} - \frac{x ^{3}}{4} + \frac{11 x ^{2}}{24} - \frac{x}{4}

求解步骤

\frac{x ( x - 1 ) ( x - 2 ) ( x - 3 )}{24}

乘开

x (x - 1) (x - 2) (x - 3) : x^{4} - 6 x^{3} + 11 x^{2} - 6 x

= \frac{x ^{4} - 6 x ^{3} + 11 x ^{2} - 6 x}{24}

使用分式法则:

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

\frac{x ^{4} - 6 x ^{3} + 11 x ^{2} - 6 x}{24} = \frac{x ^{4}}{24} - \frac{6 x ^{3}}{24} + \frac{11 x ^{2}}{24} - \frac{6 x}{24}

= \frac{x ^{4}}{24} - \frac{6 x ^{3}}{24} + \frac{11 x ^{2}}{24} - \frac{6 x}{24}

消掉

\frac{6 x ^{3}}{24} : \frac{x ^{3}}{4}

= \frac{x ^{4}}{24} - \frac{x ^{3}}{4} + \frac{11 x ^{2}}{24} - \frac{6 x}{24}

消掉

\frac{6 x}{24} : \frac{x}{4}

= \frac{x ^{4}}{24} - \frac{x ^{3}}{4} + \frac{11 x ^{2}}{24} - \frac{x}{4}

e x p an d lo g_{10} (2) (x + 2)

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

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

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

s im pl i f y (2 x - 3)^{2}