展开 (4,2,-3,5,-1)left(2,6,-1,-4,8)

解答

展开

(4, 2, - 3, 5, - 1) l e f t (2, 6, - 1, - 4, 8)

4 e l f t, 2, 2, 6, - 4 e l f t, 2, 1, - 4 e l f t, 2, 4, 8 - 3 e l f t, 5, 2, 6, 3, 5, 1, 3, 5, 4, 8 - 1 \cdot 2 e l f t, 6, 1 \cdot 1, 1 \cdot 4, 8

求解步骤

(4, 2, - 3, 5, - 1) l e f t (2, 6, - 1, - 4, 8)

= e (4 \cdot 2,, - 3 \cdot 5,, - 1) (2 \cdot 6,, - 1, - 4 \cdot 8,) l f t

乘开

(4, 2, - 3, 5, - 1) (2, 6, - 1, - 4, 8) : 4, 2, 2, 6, - 4, 2, 1, - 4, 2, 4, 8 - 3, 5, 2, 6, 3, 5, 1, 3, 5, 4, 8 - 1 \cdot 2, 6, 1 \cdot 1, 1 \cdot 4, 8

= l e f t (4, 2, 2, 6, - 4, 2, 1, - 4, 2, 4, 8 - 3, 5, 2, 6, 3, 5, 1, 3, 5, 4, 8 - 1 \cdot 2, 6, 1 \cdot 1, 1 \cdot 4, 8)

乘开

l e f t (4, 2, 2, 6, - 4, 2, 1, - 4, 2, 4, 8 - 3, 5, 2, 6, 3, 5, 1, 3, 5, 4, 8 - 1 \cdot 2, 6, 1 \cdot 1, 1 \cdot 4, 8) : 4 e l f t, 2, 2, 6, - 4 e l f t, 2, 1, - 4 e l f t, 2, 4, 8 - 3 e l f t, 5, 2, 6, 3, 5, 1, 3, 5, 4, 8 - 1 \cdot 2 e l f t, 6, 1 \cdot 1, 1 \cdot 4, 8

= 4 e l f t, 2, 2, 6, - 4 e l f t, 2, 1, - 4 e l f t, 2, 4, 8 - 3 e l f t, 5, 2, 6, 3, 5, 1, 3, 5, 4, 8 - 1 \cdot 2 e l f t, 6, 1 \cdot 1, 1 \cdot 4, 8

s im pl i f y (2 x - 5)^{2}

e x p an d (- 1, - 5) y

e x p an d \frac{2 - 4 e ^{- π s}}{( s ^{2} - 6 ) ( s ^{2} + 1 )} + \frac{s}{s ^{2} - 6}

\frac{\partial}{\partial x} ((z) (x^{2} y + y z^{2} - x z))

e x p an d f^{'} (x) (1 - 3 x^{2})^{3}