若x^2+2x+y^2-6y+10=0,求x.y的值

2025-06-27 19:36:52
推荐回答(1个)
回答1:

x^2+2x+y^2-6y+10=0
(x+1)²+(y-3)² = 0
∵(x+1)²≥0,(y-3)² ≥ 0
∴(x+1)²=0,(y-3)² = 0
∴x=-1,y=3
x*y = (-1)*3 = -3