展开 ((2x+3)^2-x^4)/2

解答

展开

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

- \frac{x ^{4}}{2} + 2 x^{2} + 6 x + \frac{9}{2}

求解步骤

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

乘开

(2 x + 3)^{2} - x^{4} : 4 x^{2} + 12 x + 9 - x^{4}

= \frac{4 x ^{2} + 12 x + 9 - x ^{4}}{2}

使用分式法则:

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

\frac{4 x ^{2} + 12 x + 9 - x ^{4}}{2} = \frac{4 x ^{2}}{2} + \frac{12 x}{2} + \frac{9}{2} - \frac{x ^{4}}{2}

= \frac{4 x ^{2}}{2} + \frac{12 x}{2} + \frac{9}{2} - \frac{x ^{4}}{2}

数字相除:

\frac{4}{2} = 2

= 2 x^{2} + \frac{12 x}{2} + \frac{9}{2} - \frac{x ^{4}}{2}

数字相除:

\frac{12}{2} = 6

= 2 x^{2} + 6 x + \frac{9}{2} - \frac{x ^{4}}{2}

改写为标准形式

= - \frac{x ^{4}}{2} + 2 x^{2} + 6 x + \frac{9}{2}

e x p an d (0 - 2 - 2) x (231) x (- 2, 1, 2)

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

e x p an d (2 π + 16) \cdot 10

e x p an d \frac{3 x ( x ^{2} - 5 )}{x ^{2} - 1}

e x p an d \frac{40 ( 4 cos ( 40 t ) - sin ( 40 t ) )}{17}