展开 ((x-1)^{4n}(x^3+8)(x-4))/(x^2-2x+2)

解答

展开

\frac{( x - 1 ) ^{4 n} ( x ^{3} + 8 ) ( x - 4 )}{x ^{2} - 2 x + 2}

\frac{x ^{4} ( x - 1 ) ^{4 n} - 4 x ^{3} ( x - 1 ) ^{4 n} + 8 x ( x - 1 ) ^{4 n} - 32 ( x - 1 ) ^{4 n}}{x ^{2} - 2 x + 2}

求解步骤

\frac{( x - 1 ) ^{4 n} ( x ^{3} + 8 ) ( x - 4 )}{x ^{2} - 2 x + 2}

乘开

(x - 1)^{4 n} (x^{3} + 8) (x - 4) : x^{4} (x - 1)^{4 n} - 4 x^{3} (x - 1)^{4 n} + 8 x (x - 1)^{4 n} - 32 (x - 1)^{4 n}

= \frac{x ^{4} ( x - 1 ) ^{4 n} - 4 x ^{3} ( x - 1 ) ^{4 n} + 8 x ( x - 1 ) ^{4 n} - 32 ( x - 1 ) ^{4 n}}{x ^{2} - 2 x + 2}

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

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

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

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

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