展开 (2n-1)(2n+!)

解答

展开

(2 n - 1) (2 n +!)

4! n^{2} - 2! n

求解步骤

(2 n - 1) (2 n +!)

= (2 n - 1) (2 +! n)

使用分配律:

a (b - c) = ab - a c

a = (2 n +!), b = 2 n, c = 1

= (2 n +!) \cdot 2 n - (2 n +!) \cdot 1

= 2 n (2 +! n) - 1 \cdot (2 +! n)

化简

2 n (2 +! n) - 1 \cdot (2 +! n) : 4! n^{2} - 2! n

= 4! n^{2} - 2! n

e x p an d \frac{( u + 3 ) ( 5 u + 1 )}{10 u ^{2} - 11 u + 1}

s im pl i f y (- 4 x + 1) (x - 5)

e x p an d \frac{e ^{k (t - a)} M}{e ^{k (t - a)} + 1}

e x p an d 10 (3 - 7)

e x p an d \frac{45 ( x - 2 ) ^{8}}{( 2 x + 1 ) ^{10}}