展开 (-2t^5-2t)/((t^2+1)^2(t^2-1)^3)

解答

展开

\frac{- 2 t ^{5} - 2 t}{( t ^{2} + 1 ) ^{2} ( t ^{2} - 1 ) ^{3}}

\frac{- 2 t ^{5} - 2 t}{t ^{10} - t ^{8} - 2 t ^{6} + 2 t ^{4} + t ^{2} - 1}

求解步骤

\frac{- 2 t ^{5} - 2 t}{( t ^{2} + 1 ) ^{2} ( t ^{2} - 1 ) ^{3}}

(t^{2} + 1)^{2} = t^{4} + 2 t^{2} + 1

= \frac{- 2 t ^{5} - 2 t}{( t ^{4} + 2 t ^{2} + 1 ) ( t ^{2} - 1 ) ^{3}}

(t^{2} - 1)^{3} = t^{6} - 3 t^{4} + 3 t^{2} - 1

= \frac{- 2 t ^{5} - 2 t}{( t ^{4} + 2 t ^{2} + 1 ) ( t ^{6} - 3 t ^{4} + 3 t ^{2} - 1 )}

乘开

(t^{4} + 2 t^{2} + 1) (t^{6} - 3 t^{4} + 3 t^{2} - 1) : t^{10} - t^{8} - 2 t^{6} + 2 t^{4} + t^{2} - 1

= \frac{- 2 t ^{5} - 2 t}{t ^{10} - t ^{8} - 2 t ^{6} + 2 t ^{4} + t ^{2} - 1}

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

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

e x p an d \frac{4 ( 285 - x )}{9}

e x p an d \frac{4 x - 9}{x ( x + 3 )}

e x p an d \frac{2 x - 2}{( 5 - 2 x + x ^{2} ) ^{4}}