换元法解方程（x²-5x）/（x+1)+（24x+24）/（x²-5）+14=0
问题描述：
换元法解方程（x²-5x）/（x+1)+（24x+24）/（x²-5）+14=0
答
设(x^2-5x)/(x+1)=tt+24/t+14=0t^2+14t+24=0(t+2)(t+12)=0t1=-2,t2=-12t1=(x^2-5x)/(x+1)=-2x^2-5x=-2x-2x^2-3x+2=0(x-1)(x-2)=0x1=1,x2=2t2=(x^2-5x)/(x+1)=-12x^2-5x=-12x-12x^2+7x+12=0(x+3)(x+4)=0x3=-3,x4=-4