已知4x^2+y^2-4x+6y+10=0,求4x^2-12xy+9y^2的值
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已知4x^2+y^2-4x+6y+10=0,求4x^2-12xy+9y^2的值
已知4x^2+y^2-4x+6y+10=0,求4x^2-12xy+9y^2的值
已知4x^2+y^2-4x+6y+10=0,求4x^2-12xy+9y^2的值
4x^2+y^2-4x+6y+10=0
(4x^2-4x+1)+(y^2+6y+9)=0
(2x-1)^2+(y+3)^2=0
x=1/2,y=-3
4x^2-12xy+9y^2
=(2x-3y)^2
=[2*1/2-3*(-3)]^2
=(1+9)^2
=100
分解因式!楼上的已经做出来了我就不再说了!
4x^2+y^2-4x+6y+10=0
(2x-1)^2+(y+3)^2=0
所以两个非负数相加只有0+0=0
即2x-1=0
y+3=0
x=1/2 y=-3
4x^2-12xy+9y^2
=(2x-3y)^2=(1+9)^2=100
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