已知x²+y²+13-4x+6y=0,求（2x-y)²-2(2x-y)(x+2y)+(x+2y)²的值

已知x²+y²+13-4x+6y=0,求（2x-y)²-2(2x-y)(x+2y)+(x+2y)²的值
已知x²+y²+13-4x+6y=0,求（2x-y)²-2(2x-y)(x+2y)+(x+2y)²的值

已知x²+y²+13-4x+6y=0,求（2x-y)²-2(2x-y)(x+2y)+(x+2y)²的值
x²+y²+13-4x+6y=0
(x²-4x+4)+(y²+6y+9)=0
(x-2)²+(y+3)²=0
平方项恒非负,两非负项之和=0,两非负项分别=0
x-2=0 x=2
y+3=0 y=-3
(2x-y)²-2(2x-y)(x+2y)+(x+2y)²
=[(2x-y)-(x+2y)]²
=(x-3y)²
=[2-3·(-3)]²
=11²
=121

x²+y²+13-4x+6y=0
（x-2）²+（y+3）²=0
x=2，y=-3
（2x-y)²-2(2x-y)(x+2y)+(x+2y)²
=【（2x-y）-（x+2y）】²
=（x-3y）²
=【2-3×（-3）】²
=11²
=121

x²+y²+13-4x+6y=0
（x-2）²+（y+3）²=0
x=2，y=-3
（2x-y)²-2(2x-y)(x+2y)+(x+2y)²
=[（2x-y)-(x+2y)]²
=(x-3y)²
=11²
=121