x-1分之1+x+1分之k-5=x平方-1分之k-11/(x-1)+(k-5)/(x+1)=(k-1)/(x²-1)(x+1+xk-k-5x+5)/(x-1)(x+1)=(k-1)/(x-1)(x+1)-4x+xk-k+6=k-1x(k-4)=2k-7x=(2k-7)/(k-4)∵增根为 x=±1∴(2k-7)/(k-4)=±12k-7=±(k-4)∴2...