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這一步是怎么算的kx²-K(x+2)=x(x+1)+6

去括號得
kx²-Kx-2k=x²+x+6
移項(xiàng)得
kx²-x²-kx-x-2k-6=0
合并同類項(xiàng)得
(k-1)x²-(k+1)x-2k-6=0
二次項(xiàng)系數(shù)不為0得
k-1≠0
移項(xiàng)得
k≠1