Question

已知a-2b=2(a≠0),求(a²-4b²)/(a²-4b²+a+2b)-a&

Answer 1

已知a-2b=2(a≠0),

(a²-4b²)/(a²-4b²+a+2b)-a²+4ab-4b²

= (a+2b)(a-2b) / (a-2b)(a+2b) + (a+2b) - (a-2b) ²

= (a+2b)(a-2b) / (a-2b+1)(a+2b) - (a-2b) ²

= (a-2b) / (a-2b+1) - (a-2b) ²

代入 a-2b=2

= 2 /( 2+1) - 2x2

= 2/3 -4

= - 10/3

=-3分之10