展开 (X(1-X)^3)/3

解答

展开

\frac{X ( 1 - X ) ^{3}}{3}

- \frac{X ^{4}}{3} + X^{3} - X^{2} + \frac{X}{3}

求解步骤

\frac{X ( 1 - X ) ^{3}}{3}

(1 - X)^{3} = 1 - 3 X + 3 X^{2} - X^{3}

= \frac{X ( - X ^{3} + 3 X ^{2} - 3 X + 1 )}{3}

乘开

X (1 - 3 X + 3 X^{2} - X^{3}) : X - 3 X^{2} + 3 X^{3} - X^{4}

= \frac{X - 3 X ^{2} + 3 X ^{3} - X ^{4}}{3}

使用分式法则:

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

\frac{X - 3 X ^{2} + 3 X ^{3} - X ^{4}}{3} = \frac{X}{3} - \frac{3 X ^{2}}{3} + \frac{3 X ^{3}}{3} - \frac{X ^{4}}{3}

= \frac{X}{3} - \frac{3 X ^{2}}{3} + \frac{3 X ^{3}}{3} - \frac{X ^{4}}{3}

数字相除:

\frac{3}{3} = 1

= \frac{X}{3} - X^{2} + X^{3} - \frac{X ^{4}}{3}

改写为标准形式

= - \frac{X ^{4}}{3} + X^{3} - X^{2} + \frac{X}{3}

e x p an d 133 \cdot 24

e x p an d - \frac{( t ^{2} + t ) x + 6 x}{2}

e x p an d a c (er + f) + a d + b

e x p an d \frac{5}{( a + 3 ) ( h + a + 3 )}

e x p an d (2 + 3 i) (4 + i)^{''}