解:配方,得:(a+1)^2+(b-3)^2=0
因为,对任意实数a,恒有a^2≥0
所以,必有(a+1)^2=0,(b-3)^2=0
所以,a=-1,b=3
(a²+2a+1)+(b²-6b+9)=0
(a+1)²+(b-3)²=0
则:a+1=0且b-3=0
得:a=-1、b=3
a²+2a+b²-6b+10=0
(a+1)²+(b-3)²=0
a+1=0
b-3=0
a=-1
b=3
其实很简单 (a+1)(a+1)+(b-3)(b-3)=0 a=-1 b=3 把那个10拆开就行
a²+2a+b²-6b+10=0
a²+2a+1+b²-6b+9=0
(a+1)²+(b-3)²=0
(a+1)²=0,(b-3)²=0
a=-1
b=3