令：y = x²+2x = (x+1)²-1 ≥ -1原方程为：y + (m²-1)/(y-2m) = 0--->y(y-2m) + (m²-1) = 0--->y²-2my+(m²-1) = 0--->[y-(m+1)][y-(m-1)] = 0--->y1=m+1，y2=m-1（1）x无解--->y的解都 展开
令：y = x²+2x = (x+1)²-1 ≥ -1原方程为：y + (m²-1)/(y-2m) = 0--->y(y-2m) + (m²-1) = 0--->y²-2my+(m²-1) = 0--->[y-(m+1)][y-(m-1)] = 0--->y1=m+1，y2=m-1（1）x无解--->y的解都小于-1--->m+1＜-1--->m＜-2（2）x恰有三解--->y的一个解恰为-1--->m=-2或m=0 m=-2时，y1=-1,y2=-3＜-1，舍去 m=0时，y2=-1,y1=1----->x1=-1，x2,3=(-1±√2)/2 收起
