已知a,b,y,x是有理数,且x-a的绝对值+(y+b)²=0

求[(a²+ay-bx+b²)/(x+y)]/[(a²+ax+by-b²)/(a+b)]
2025-12-17 02:40:27
推荐回答(1个)
回答1:

绝对值和平方都大于等于0
相加等于0则都等于0
所以x-a=0,y+b=0
x=a,y=-b
[(a²+ay-bx+b²)/(x+y)]/[(a²+ax+by-b²)/(a+b)]
=[(a²-2ab+b²)/(a-b)]/[(a²+a²-b²-b²)/(a+b)]
=[(a-b)²/(a-b)]/[2(a+b)(a-b)/(a+b)]
=(a-b)/2(a-b)
=1/2