已知xy满足x2+y2-6x+2y+10=0,求立方根号x2-y2的值
问题描述：
已知xy满足x2+y2-6x+2y+10=0,求立方根号x2-y2的值
答
x2+y2-6x+2y+10=0
(x-3)^2+(y+1)^2=0
x=3 y=-1
答
x^2+y^2-6x+2y+10=0
x^2-6x+9+y^2+2y+1=0
(x-3)^2+(y+1)^2=0
x-3=0 x=3
y+1=0 y=-1
x^2-y^2
=3^2-(-1)^2
=9-1
=8
(x^2-y^2)^(1/3)=8^(1/3)=2
答
条件变换：
(x-3)^2+(y+1)^2=0
即：
y+1=0
x-3=0
所以：
立方根号x2-y2=2
答
x2+y2-6x+2y+10=0
(x-3)^2 + (y+1)^2 = 0
=> x = 3, y = -1
(x^2 - y^2)^(1/3)
= ( 8)^(1/3)
= 2