若多项式x²-2(kx-1)y+xy-x+y²不含xy项,求k的值.
人气:438 ℃ 时间:2019-10-24 03:11:33
解答
不含xy项,则合并同类项后xy的系数为0,则
x²-2(kx-1)y+xy-x+y²
=x²-2kxy+2y+xy-x+y²
=x²+(1-2k)xy+2y-x+y²
于是
1-2k=0
k=1/2
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