已知x^2+y^2-6x-4y+13=0,求(2x-y)^2-2(2x-y)(x+2y)+(x+2y)^2的值

x^2+y^2-6x-4y+13=0
（x-3）^2+(y-2)^2=0
x-3=0 x=3
y-2=0 y=2
(2x-y)^2-2(2x-y)(x+2y)+(x+2y)^2
=[(2x-y)-(x+2y)]^2
=(x-3y)^2
x=3,y=2
=(3-2*3)^2
=9
希望能够帮助你!
再问： y=1/3x-1,求1/3x^2-2xy+3y^2-2