换元法解方程（x²-5x）/（x+1)+（24x+24）/（x²-5）+14=0

换元法解方程（x²-5x）/（x+1)+（24x+24）/（x²-5）+14=0
换元法解方程（x²-5x）/（x+1)+（24x+24）/（x²-5）+14=0

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