已知a,b,y,x是有理数，且x-a的绝对值+（y+b）²=0

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

x-a的绝对值+（y+b）²=0
x-a=0,y+b=0,
x=a,y=-b代入[(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）
=1/2

∵|x-a|+（y+b）²=0，∴x=a，y=-b。
∴(a²+ay-bx+b²）/（x+y)
=(a²-ab-ab+b²）/（a-b)
=a-b，

(a²+ax+by-b²)/(a+b)
=(a²+aa-bb-b²)/(a+b)
=2（a-b）

∴原式=(a-b)/2(a-b)=1/2.

x-a的绝对值+（y+b）²=0
x=a,y=-b
[(a²+ay-bx+b²）/（x+y)]/[(a²+ax+by-b²)/(a+b)]
=[(a^2-2ab+b^2)/(a-b)]/[2(a^2-b^2)/(a+b)]
=(a-b)/2(a-b)
=1/2