若多项式x²-2(kx-1)y+xy-x+y²不含xy项,求k的值.
问题描述：
若多项式x²-2(kx-1)y+xy-x+y²不含xy项,求k的值.
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