展开 (3(x+2)^4)/4-(5(x+2)^3)/3

解答

展开

\frac{3 ( x + 2 ) ^{4}}{4} - \frac{5 ( x + 2 ) ^{3}}{3}

\frac{3 x ^{4}}{4} + \frac{13 x ^{3}}{3} + 8 x^{2} + 4 x - \frac{4}{3}

求解步骤

\frac{3 ( x + 2 ) ^{4}}{4} - \frac{5 ( x + 2 ) ^{3}}{3}

\frac{3 ( x + 2 ) ^{4}}{4} = \frac{3 x ^{4} + 24 x ^{3} + 72 x ^{2} + 96 x + 48}{4}

\frac{5 ( x + 2 ) ^{3}}{3} = \frac{5 x ^{3} + 30 x ^{2} + 60 x + 40}{3}

= \frac{3 x ^{4} + 24 x ^{3} + 72 x ^{2} + 96 x + 48}{4} - \frac{5 x ^{3} + 30 x ^{2} + 60 x + 40}{3}

4, 3

的最小公倍数:

12

根据最小公倍数调整分式

= \frac{( 3 x ^{4} + 24 x ^{3} + 72 x ^{2} + 96 x + 48 ) \cdot 3}{12} - \frac{( 5 x ^{3} + 30 x ^{2} + 60 x + 40 ) \cdot 4}{12}

因为分母相等，所以合并分式:

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

= \frac{( 3 x ^{4} + 24 x ^{3} + 72 x ^{2} + 96 x + 48 ) \cdot 3 - ( 5 x ^{3} + 30 x ^{2} + 60 x + 40 ) \cdot 4}{12}

乘开

(3 x^{4} + 24 x^{3} + 72 x^{2} + 96 x + 48) \cdot 3 - (5 x^{3} + 30 x^{2} + 60 x + 40) \cdot 4 : 9 x^{4} + 52 x^{3} + 96 x^{2} + 48 x - 16

= \frac{9 x ^{4} + 52 x ^{3} + 96 x ^{2} + 48 x - 16}{12}

使用分式法则:

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

\frac{9 x ^{4} + 52 x ^{3} + 96 x ^{2} + 48 x - 16}{12} = \frac{9 x ^{4}}{12} + \frac{52 x ^{3}}{12} + \frac{96 x ^{2}}{12} + \frac{48 x}{12} - \frac{16}{12}

= \frac{9 x ^{4}}{12} + \frac{52 x ^{3}}{12} + \frac{96 x ^{2}}{12} + \frac{48 x}{12} - \frac{16}{12}

\frac{9 x ^{4}}{12} = \frac{3 x ^{4}}{4}

\frac{52 x ^{3}}{12} = \frac{13 x ^{3}}{3}

\frac{96 x ^{2}}{12} = 8 x^{2}

\frac{48 x}{12} = 4 x

\frac{16}{12} = \frac{4}{3}

= \frac{3 x ^{4}}{4} + \frac{13 x ^{3}}{3} + 8 x^{2} + 4 x - \frac{4}{3}

e x p an d \frac{x ^{4} - 6 x ^{3} + 12 x ^{2}}{( x ^{2} - 3 x + 4 ) ^{2}}

e x p an d n gu l a r (220, - 120)

\frac{d ^{2}}{d x ^{2}} ((y) (e^{2 x} - k x))

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

e x p an d \frac{6}{( s - 1 ) ^{2}}