Question

有一道题a²+b²-4a-4b+2ab+3<0 a和b为正整数,求a+b是多少

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

Answer 1

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