方程化为：1/(x+1)+2k/(x-4)=(3k+1)/[(x-4)(x+1)]两边同时乘以(x-4)(x+1)得：x-4+2k(x+1)=(3k+1)x(2k+1)=k+5若2k+1=0, 即k=-1/2时,方程无解,原方程也无解若2k+1≠0, 则x=(k+5)/(2k+1), 若原方程无解,则此根为增根,有...