换元法：令(x^2-5x)/(x+1)=t,则(x+1)/(x^2-5x)=1/t,所以原方程换元后为t+24/t+14=0,方程两边同时乘以t得t^2+14t+24=0解得t=-2或t=-12即(x^2-5x)/(x+1)=-2或(x^2-5x)/(x+1)=-12,解(x^2-5x)/(x+1)=-2得:x=1或x=2 解(x^2-5x)/(x+1)=-12得:x=-3或x=-4这题挺难得,仔细看看吧.