初一 数学 数学 请详细解答,谢谢! (7 22:14:4)

x2+4x+y2-6y+13=0
(x+2)^2+(y-3)^2=0
x=-2 y=3
(x+2y)(x-2y)
=x^2-4y^2
=4-36
=-32

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)(x-2y)
=(-2+6)(-2-6)
=-32

x2+4x+y2-6y+13=0
x^2+4x+4+y^2-6y+9=0
(x+2)^2+(y-3)^2=0
则
x+2=0,y-3=0,
x=-2,y=3,
从而
(x+2y)(x-2y)
=x^2-(2y)^2
=x^2-4y^2
=(-2)^2-4*3^2
=4-36
=-32

x2+4x+y2-6y+13
=x2+4x+4+y2-6y+9
=（x+2）^2+(y-3)^2=0
则x=-2 y=3
(x+2y)(x-2y）=（-2+2*3)*(-2-2*3)=-32

∵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)^2≥0,(y-3)^2≥0

∴x+2=0,y-3=0

∴x=-2,y=3

∴(x+2y)(x-2y)=(-2+6)(-2-6)=4×(-8)=-32

x2+4x+y2-6y+13=0
(x2+4x+4)+(y2-6y+9)=0
所以(x+2)^2+(y-3)^2=0 所以x+2=0,y-3=0
所以x=-2 y=3
则(x+2y)(x-2y)
=x^2-4y^2
把x=-2 y=3 带入x^2-4y^2
则=