1.[(x²+1)/x]²+(x²+1)/x=2
设(x²+1)/x=y,则原方程化为y²+y-2=0,解得y1=-2 y2=1
即(x²+1)/x=-2或(x²+1)/x=1
解得x1=x2=-1
2.(x²-5x)/(x+1)+24(x+1)/(x²-5x)+14=0
设(x²-5x)/(x+1)=y,则原方程化为y+24/y+14=0,即y²+14y+24=0,解得y1=-2 y2=-12
即(x²-5x)/(x+1)=-2或(x²-5x)/(x+1)=-12
整理得x²-3x+2=0或x²+7x+12=0
解得x1=1 x2=2 x3=-3 x4=-4