设a是实数,解关于x的方程:(a-1)x2-2ax+a=0

1）a=1时,方程成为-2x+1=0--->x=1/2.
2)a1时,两边同除以a-1,得到 x^2-2ax/(a-1)+a/(a-1)=0
--->x^2-2ax/(a-1)+a^2/(a-1)^2=-a/(a-1)+a^2/(a-1)^2
--->[x-a/(a-1)]^2=a/(a-1)^2
--->x-a/(a-1)=+ -√a/(a-1)
--->x=(a+ -√a)/(a-1).