已知x²+y²-2x-6y+10=0,求4(x²+y)(x²-y)-(2x
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已知x²+y²-2x-6y+10=0,求4(x²+y)(x²-y)-(2x²-y)²
![已知x²+y²-2x-6y+10=0,求4(x²+y)(x²-y)-(2x](/uploads/image/z/16137581-5-1.jpg?t=%E5%B7%B2%E7%9F%A5x%26%23178%3B%2By%26%23178%3B-2x-6y%2B10%3D0%2C%E6%B1%824%28x%26%23178%3B%2By%29%28x%26%23178%3B-y%29-%282x%26%231)
即(x²-2x+1)+(y²-6y+9)=0
(x-1)²+(y-3)²=0
所以x-1=0,y-3=0
x=1,y=3
所以原式=4x⁴-4y²-4x⁴+4x²y-y²
=4x²y-5y²
=4*1*3-5*9
=-33
(x-1)²+(y-3)²=0
所以x-1=0,y-3=0
x=1,y=3
所以原式=4x⁴-4y²-4x⁴+4x²y-y²
=4x²y-5y²
=4*1*3-5*9
=-33
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