展开 (1,2,-2)left(1,-1,-1)

解答

展开

(1, 2, - 2) l e f t (1, - 1, - 1)

1 e l f t, 2, 1, - 1 e l f t, 2, 1, - 1 e l f t, 2, 1 - 2 \cdot 1 e l f t, 2 \cdot 1, 2 \cdot 1

求解步骤

(1, 2, - 2) l e f t (1, - 1, - 1)

= e (1 \cdot 2,, - 2) (1, - 1, - 1) l f t

乘开

(1, 2, - 2) (1, - 1, - 1) : 1, 2, 1, - 1, 2, 1, - 1, 2, 1 - 2 \cdot 1, 2 \cdot 1, 2 \cdot 1

= l e f t (1, 2, 1, - 1, 2, 1, - 1, 2, 1 - 2 \cdot 1, 2 \cdot 1, 2 \cdot 1)

乘开

l e f t (1, 2, 1, - 1, 2, 1, - 1, 2, 1 - 2 \cdot 1, 2 \cdot 1, 2 \cdot 1) : 1 e l f t, 2, 1, - 1 e l f t, 2, 1, - 1 e l f t, 2, 1 - 2 \cdot 1 e l f t, 2 \cdot 1, 2 \cdot 1

= 1 e l f t, 2, 1, - 1 e l f t, 2, 1, - 1 e l f t, 2, 1 - 2 \cdot 1 e l f t, 2 \cdot 1, 2 \cdot 1

s im pl i f y \frac{x + 3}{( 3 x - 2 ) ^{\frac{2}{3}}} + \frac{3 3 x - 2}{2 x}

e x p an d (1 + sin (x))^{2} - cos^{2} (x)

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

e x p an d \frac{3 ( 9 - y )}{2}

e x p an d \frac{cos ^{2} ( x )}{( 1 + sin ( x ) ) ^{2}}