Question

先化简，再求值1-(a-b/a+2b)除以(a²-b²/a²+4ab+4b²),其中a=2,b=1

RT、

Answer 1

原式=1-[(a-b)/(a+2b)]÷[(a+b)(a-b)/(a+2b)²]
=1-[(a-b)/(a+2b)]×[(a+2b)²/(a+b)(a-b)]
=1-(a+2b)/(a+b)
=(a+b-a-2b)/(a+b)
=-b/(a+b)
=-1/(2+1)
=-1/3

Answer 2

解：1-(a-b/a+2b)除以(a²-b²/a²+4ab+4b²)
=1-[(a-b)/(a+2b)][(a²+4ab+4b²)/(a²-b²)]
=1-[(a-b)/(a+2b)][(a+b)^2/(a-b)(a+b)]]
=1-(a+2b)/(a+b)
=-b/(a+b)
=-1/3