x^2+y^2+4x-6y+13=0
x^y=?
(x^2+4x+4)+(y^2-6y+9)=0
(x+2)^2+(y-3)^2=0
x+2=0
y-3=0
x=-2,
y=3
x^y=(-2)^3=-8
第二题
由式a^2+b^2+c^2=ab+bc+ac 乘以2得
2a^2+2b^2+2c^2=2ab+2bc+2ac
移向得(a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)=0
(a-b)^2+(b-c)^2+(a-c)^2=0
(a-b)^2=0
(b-c)^2=0
(a-c)^2=0
得a=b=c
代入a+2b+3c=12得a=b=c=2
（a+b）^2+c^3=24