展开 re(z-1)

解答

展开

re (z - 1)

erz - er

求解步骤

re (z - 1)

= er (z - 1)

使用分配律:

a (b - c) = ab - a c

a = re, b = z, c = 1

= rez - re 1

= erz - 1 er

乘以:

1 e = e

= erz - er

e x p an d \frac{- \frac{5}{4} z}{z - 0.3} + \frac{\frac{9}{4} z}{z - 0.7}

e x p an d 9 (x + Δ x)^{2} - 9 (x + Δ x)

s im pl i f y 5 (25 - x^{2} + 5) 25 - x^{2}

e x p an d \frac{1}{( 3 - x ) ^{3}}

e x p an d \frac{- 3 x ^{2} - 3 x + 27}{( x + 2 ) ( 2 x ^{2} + 3 x - 9 )}