已知x²-xy-2y²=x-2y,且x-2y≠0,则:(1)x+y=_______(2)若题中x、y满足等式x²+y²-9x+5y+26=0,请确定x、y的值.
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![已知x²-xy-2y²=x-2y,且x-2y≠0,则:(1)x+y=_______(2)若题中x、y满足等式x²+y²-9x+5y+26=0,请确定x、y的值.](/uploads/image/z/1709966-38-6.jpg?t=%E5%B7%B2%E7%9F%A5x%26%23178%3B-xy-2y%26%23178%3B%3Dx-2y%2C%E4%B8%94x-2y%E2%89%A00%2C%E5%88%99%3A%281%29x%2By%3D_______%282%29%E8%8B%A5%E9%A2%98%E4%B8%ADx%E3%80%81y%E6%BB%A1%E8%B6%B3%E7%AD%89%E5%BC%8Fx%26%23178%3B%2By%26%23178%3B-9x%2B5y%2B26%3D0%2C%E8%AF%B7%E7%A1%AE%E5%AE%9Ax%E3%80%81y%E7%9A%84%E5%80%BC.)
已知x²-xy-2y²=x-2y,且x-2y≠0,则:(1)x+y=_______(2)若题中x、y满足等式x²+y²-9x+5y+26=0,请确定x、y的值.
已知x²-xy-2y²=x-2y,且x-2y≠0,则:(1)x+y=_______
(2)若题中x、y满足等式x²+y²-9x+5y+26=0,请确定x、y的值.
已知x²-xy-2y²=x-2y,且x-2y≠0,则:(1)x+y=_______(2)若题中x、y满足等式x²+y²-9x+5y+26=0,请确定x、y的值.
(1) 将等式变形为 (x+y)(x-2y)=x-2y ,因x-2y≠0,两边除以x-2y得 x+y = 1
(2) 将x=1-y代入方程,得
(1-y)²-9(x+y)+14y+26=0
整理得 y²+6y+9=0 即 (y+3)²=0
故 y= - 3 ,x=1-y = 4
(1)原式可化为:(x-2y)²+3xy-6y²=x-2y
(x-2y)²+3y(x-2y)=x-2y
∵x-2y≠0
∴x-2y+3y=1
x+y=1
(2)原式可化为:(x+y)²-2xy-9x+5y=-26
把...
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(1)原式可化为:(x-2y)²+3xy-6y²=x-2y
(x-2y)²+3y(x-2y)=x-2y
∵x-2y≠0
∴x-2y+3y=1
x+y=1
(2)原式可化为:(x+y)²-2xy-9x+5y=-26
把x+y=1代入式中,得:5y-2xy-9x=-27
∵x+y=1
∴y=1-x
把y=1-x代入原式中,化简后得x²-8x=-16
解得: x=4
y=-3
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