Question

已知x、y满足x2+y2-4x+6y+13=0 求代数式（3x+y)2-3(3x-y)(x+y)-(x-3y)(x+3y)

已知x、y满足x^2+y^2-4x+6y+13=0 求代数式（3x+y)^2-3(3x-y)(x+y)-(x-3y)(x+3y)的值
急！

Answer 1

x2+y2-4x+6y+13=0
x²-4x+4+y²+6y+9=0
(x-2)²+(y+3)²=0
∴x=2 y=-3
（3x+y)2-3(3x-y)(x+y)-(x-3y)(x+3y)
=9x²+6xy+y²-9x²-6xy+3y²-x²+9y²
=13y²-x²
=13*(-3)²-2²
=117-4
=113

Answer 2

x2+y2-4x+6y+13=0
(x-2)^2+(y-3)^2=0
x-2=0 y-3=0
x=2 y=3
（3x+y)2-3(3x-y)(x+y)-(x-3y)(x+3y)
=9x^2+6xy+y^2-9x^2-6xy+3y^2-x^2+9y^2
=-x^2+13y^2
=-4+13*9
=113

Answer 3

x2+y2-4x+6y+13=0
(x-2)^2+(y+3)^2 = 0
所以，x = 2 ，y = -3
（3x+y)2-3(3x-y)(x+y)-(x-3y)(x+3y)=113