展开 (((x+4)(6a+x-4)))/6

解答

展开

\frac{( ( x + 4 ) ( 6 a + x - 4 ) )}{6}

a x + \frac{x ^{2}}{6} + 4 a - \frac{8}{3}

求解步骤

\frac{( x + 4 ) ( 6 a + x - 4 )}{6}

乘开

(x + 4) (6 a + x - 4) : 6 a x + x^{2} + 24 a - 16

= \frac{6 a x + x ^{2} + 24 a - 16}{6}

使用分式法则:

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

\frac{6 a x + x ^{2} + 24 a - 16}{6} = \frac{6 a x}{6} + \frac{x ^{2}}{6} + \frac{24 a}{6} - \frac{16}{6}

= \frac{6 a x}{6} + \frac{x ^{2}}{6} + \frac{24 a}{6} - \frac{16}{6}

数字相除:

\frac{6}{6} = 1

= a x + \frac{x ^{2}}{6} + \frac{24 a}{6} - \frac{16}{6}

数字相除:

\frac{24}{6} = 4

= a x + \frac{x ^{2}}{6} + 4 a - \frac{16}{6}

消掉

\frac{16}{6} : \frac{8}{3}

= a x + \frac{x ^{2}}{6} + 4 a - \frac{8}{3}

\frac{\partial}{\partial x} ((z) (z^{3} - 3 x yz))

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

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

e x p an d 3 (e - 1)

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