令 x^2-2x=t则 2x^2-4x=2t原方程就是：t - 3/(t-1)=3(t-3)(t-1)=3t^2-4t=0t1=0,t2=4 ,都有意义当t1=0时,x^2-2x=0,解得 x1=0,x2=2当t2=4时,x^2-2x=4,有(x-1)^2=5 解得 x3=1+√5 ,x4=1-√5即方程有上面的4个解....