已知X^2+4X+Y^2-6Y+13=0.求(X-2Y)除以(X^2+Y^2)的值
问题描述:
已知X^2+4X+Y^2-6Y+13=0.求(X-2Y)除以(X^2+Y^2)的值
答
已知X^2+4X+Y^2-6Y+13=0。求(X-2Y)除以(X^2+Y^2)的值
x^2-4x+y^2-6y+13=0
(x^2-4x+4)+(y^2-6y+9)=0
(x-2)^2+(y-3)^2=0
x=2
y=3
X-2Y=2-6=-4
X^2+Y^2=4+9=13
(X-2Y)除以(X^2+Y^2)的值 =-4/13
答
X^2+4X+Y^2-6Y+13=(x+2)^2+(y-3)^2=0
所以x=-2,y=3,可解题
答
X^2+4X+Y^2-6Y+13=0可化为
(X+2)^2+(y-3)^2=0
所以X=-2
Y=3
(X-2Y)除以(X^2+Y^2)=-8/13