1,(x-y+3)^2>=0,√2-y>=0,所以只有(x-y+3)^2=0,√2-y=0
于是x-y+3=0,2-y=0,解得y=2,x=-1
x+y=1;
2,展开得a^2+2a-3+a^2-2a=2a^2-3
(x-y+3)^2+√2-y=0
则,x-y+3=0且2-y=0
解得x=-1 ,y=2
x+y=-1+2=1
(a+3)(a-1)+a(a-2)
=a²+2a-3+a²-2a
=2a²-3
(1)x=-1 y=2,x+y=1;
(2)展开得2a²-1
(1) -3+√2
(2)2a²-3