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(化成完全平方来解）

Answer 1

因为x²迅滑+y²-6x+4y+13=0
(x²-6x+9)+(y²+4y+4)=0
(x-3)²+(y+2)²=0
而(x-3)²≥0，且(y+2)²≥亩帆腊0
所以x-3=0，且y+2=0
所以x=3，y=-2
那么(2x-y)²-2(2x-y)(x+2y)+(x+2y)²=[(2x-y)-(x+2y)]²
=(2x-y-x-2y)²
=(x-3y)²
=(3+6)²轿裤
=81

Answer 2

即(x²-6x+9)+(y²乎猜+4y+4)=0
(x-3)²+(y+2)²=0
x-3=y+2=0
x=3,y=-2

原式=4x²岁拦型-4xy+y²-4x²-6xy+4y²+x²+4xy+4y²衡氏
=x²-6xy+9y²
=9+36+36
=81

Answer 3

(2x-y)²-2(2x-y)(x+2y)+(x+2y)²=[(2x-y)-(x+2y)]²
=(2x-y-x-2y)²
=(x-3y)²
=(3+6)²
=81