若4x^2-4x+9y^2-12y+5=0,求(x-3y)^2-(x+3y)^2
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若4x^2-4x+9y^2-12y+5=0,求(x-3y)^2-(x+3y)^2
![若4x^2-4x+9y^2-12y+5=0,求(x-3y)^2-(x+3y)^2](/uploads/image/z/16465421-29-1.jpg?t=%E8%8B%A54x%5E2-4x%2B9y%5E2-12y%2B5%3D0%2C%E6%B1%82%EF%BC%88x-3y%EF%BC%89%5E2-%28x%2B3y%29%5E2)
4x^2-4x+9y^2-12y+5=0
4x^2-4x+1+9y^2-12y+4=0
(2x-1)^2+(3y-2)^2=0
所以
2x-1=0
x=1/2
3y-2=0
y=2/3
(x-3y)^2-(x+3y)^2
=(x-3y+x+3y)(x-3y-x-3y)
=2x*(-6y)
=-12xy
=-12*1/2*2/3
=-4
4x^2-4x+1+9y^2-12y+4=0
(2x-1)^2+(3y-2)^2=0
所以
2x-1=0
x=1/2
3y-2=0
y=2/3
(x-3y)^2-(x+3y)^2
=(x-3y+x+3y)(x-3y-x-3y)
=2x*(-6y)
=-12xy
=-12*1/2*2/3
=-4
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