Question

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

Answer 1

(x^2-6x+9)+(y^2+4y+4)=0

(x-3)^2+(y+2)^2=0

所以x-3=0,y+2=0

x=3,y=-2

原式=4x^2-4xy+y^2-4x^2-6xy+y^2+x^2+4xy+4y^2

=x^2-6xy+6y^2

=9+36+24

=69