x的平方+4x+y的平方-6y+13=0,则（x+2y)(x-2y)的值是多少?
问题描述：
x的平方+4x+y的平方-6y+13=0,则（x+2y)(x-2y)的值是多少?
答
x²+4x+y²-6y+13
=(x+2)²+(y-3)²
=0
x=-2, y=3
(x+2y)(x-2y)
=x²-4y²
=4-4*9=-32
答
x的平方+4x+y的平方-6y+13=0
(x+2)^2+(y-3)^2=0
x=-2
y=3
（x+2y)(x-2y)=x^2-4y^2=-32
答
x的平方+4x+y的平方-6y+13
=(x^2+4x+4)+(y^2-6y+9)
=(x+2)^2+(y-3)^2
=0
x=-2,y=3
（x+2y)(x-2y)
=x^2-4y^2
=4-4*9
=-32
答
x^2+4x+y^2-6y+13=0 (x+2)^2+(y-9)^2=0 x=-2 y=9
(x+2y)(x-2y)=x^2-4y^2=4-4*81=-320
答
(x^2+4x+4)+(y^2-6y+9)=0
(x+2)^2+(y-3)^2=0
平方大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立.
所以两个都等于0
所以x+2=0,y-3=0
x=-2,y=3
(x+2y)(x-2y)=4*(-8)=-32