已知A.B.X.Y是有理数.且(X-A)的绝对值+(Y+B)^2=0

求(A^2+AY-BX+B^2)/(X+Y)除以(A^2+AX+BY-B^2)/(A+B)的值
2025-12-14 11:31:31
推荐回答(2个)
回答1:

|x-a|+(y-b)²=2
那么有x-a=0即x=a,y+b=0即y=-b
[(a²+ay-bx+b²)/(x+y)]/[(a²+ax+by-b²)/(a+b)]
=[(a²-ab-ab+b²)/(a-b)]/[(a²+a²-b²-b²)/(a+b)]
=[(a²-2ab+b²)/(a-b)]/[2(a²-b²)/(a+b)]
=[(a-b)²/(a-b)]/[2(a+b)(a-b)/(a+b)]
=(a-b)/[2(a-b)]
=1/2

回答2:

根据条件可知X =A,Y=-B,所以下面式子的值为(A^2+B^2)/2(A-B)^2