原方程为:x^2+[3x/(x-3)]^2=16 对方程左边进行配方:x^2+2x*3x/(x-3)+[3x/(x-3)]^2=16+2x*3x/(x-3) 合并移项得到:[x+ 3x/(x-3)]^2- 6x^2/(x-3)-16=0 [ x^2/(x-3) ]^2- 6x^2/(x-3)-16=0 把x^2/(x-3) 看作一整体,该方程...