已知x^2+y^2+4x-6y+13=0,x,y都为有理数,求x^y的值

2024-12-28 14:36:41
推荐回答(4个)
回答1:

式子分解为:
x^2+4x+4+y^2-6y+9=0

(x+2)^2+(y-3)^2=0
所以
x = -2 ,y = 3
因此
x^y = (-2)^3 = - 8

回答2:

x^2+y^2+4x-6y+13=(x+2)^+(y-3)^2=0,因为xy均为有理数,所以x=-2,y=3
x^y==-8

回答3:

x^2+y^2+4x-6y+13=0
x^2+4x+4+y^2-6y+9=0
(x+2)²+(y-3)²=0
x=-2,y=3
x^y=-8

回答4:

x^2+y^2+4x-6y+13=0
(x+2)^2+(y-3)^2=0
x+2=0 x=-2
y-3=0 y=3
x^y=(-2)^3=-8