一道初一数学题（要过程要过程要过程要过程）

因为(x+y-3)^2≥0,(x-y+5)^2≥0
(x+y-3)^2+(x-y+5)^2=0
所以(x+y-3)^2=0,(x-y+5)^2=0
所以x+y-3=0
x-y+5=0
解此方程得x=-1
y=4
所以x^2-y^2
=(-1)^2-4^2
=1-16
=-15

(x+y-3)^2+(x-y+5)^2=0
x+y-3=0--->x+y=3
x-y+5=0--->x-y=-5

x^2-y^2=(x-y)(x+y)=-5*3=-15

因为(x+y-3)^2+(x-y+5)^2=0
=〉x+y-3=0 x-y+5=0
=〉x+y=3 x-y=-5
=〉x=-1 y=4
=〉x^2-y^2=-10

(x+y-3)^2=0，(x-y+5)^2=0，
x+y=3,x-y=-5
x=-1,y=4
x^2-y^2=-15