kx²-K(x+2)=x(x+1)+6
kx²-Kx-2k=x²+x+6
kx²-x²-kx-x-2k-6=0
(k-1)x²-(k+1)x-2k-6=0
k-1≠0
k≠1(k-1)x²-(k+1)x-2k-6=0这一步是怎么算的kx²-K(x+2)=x(x+1)+6

去括号得
kx²-Kx-2k=x²+x+6
移项得
kx²-x²-kx-x-2k-6=0
合并同类项得
(k-1)x²-(k+1)x-2k-6=0
二次项系数不为0得
k-1≠0
移项得
k≠1