当a^+4b^-4a+4b+5=0时,求(a/a-b——a^/a^-2ab+b^)/(a/a+b_a^/a^-b^)的值
3个回答

a²+4b²-4a+4b+5=0

(a²-4a+4）+（4b²+4b+1)=0

(a-2)²+（2b+1）²=0

两个平方数的和为0,这两个数都是0

所以

a-2=0

2b+1=0

a=2,b=-1/2

你后面的式子,看着眼晕,猜着做做吧,如果不对,你自己把a,b代入吧

[a/(a-b)-a²/(a²-2ab+b²)]/[a/(a+b)-a²/(a²-b²)]

=[a(a-b)/(a-b)²-a²/(a-b)²]/[a(a-b)/(a²-b²)-a²(a²-b²)]

=[(a²-ab-a²)/(a-b)²]/[(a²-ab-a²)/(a²-b²)]

=[-ab/(a-b)²]/[-ab/(a²-b²)]

=(a²-b²)/(a-b)²

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

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

=(3/2)/(5/2)

=3/5

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