展开 ((-sin^2(u)+3)sqrt(-sin^2(u)+3))/9

解答

展开

\frac{( - sin ^{2} ( u ) + 3 ) - sin ^{2} ( u ) + 3}{9}

- \frac{sin ^{2} ( u ) - sin ^{2} ( u ) + 3}{9} + \frac{- sin ^{2} ( u ) + 3}{3}

求解步骤

\frac{( - sin ^{2} ( u ) + 3 ) - sin ^{2} ( u ) + 3}{9}

乘开

(- sin^{2} (u) + 3) - sin^{2} (u) + 3 : - sin^{2} (u) - sin^{2} (u) + 3 + 3 - sin^{2} (u) + 3

= \frac{- sin ^{2} ( u ) - sin ^{2} ( u ) + 3 + 3 - sin ^{2} ( u ) + 3}{9}

使用分式法则:

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

\frac{- sin ^{2} ( u ) - sin ^{2} ( u ) + 3 + 3 - sin ^{2} ( u ) + 3}{9} = - \frac{sin ^{2} ( u ) - sin ^{2} ( u ) + 3}{9} + \frac{3 - sin ^{2} ( u ) + 3}{9}

= - \frac{sin ^{2} ( u ) - sin ^{2} ( u ) + 3}{9} + \frac{3 - sin ^{2} ( u ) + 3}{9}

消掉

\frac{3 - sin ^{2} ( u ) + 3}{9} : \frac{- sin ^{2} ( u ) + 3}{3}

= - \frac{sin ^{2} ( u ) - sin ^{2} ( u ) + 3}{9} + \frac{- sin ^{2} ( u ) + 3}{3}

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

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

e x p an d 5 + (5 (1 - e)) e^{- x}

e x p an d (- e^{t} cos (t) - sin (t))^{2}

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