`f(x).g(x)+x^2[1-3.g(x)] = 5/2`
`⇔f(x).g(x)+x^2-3x^2.g(x)=5/2` $(1)$
Thay $f(x)=3x^2-x+1;g(x)=x-1$ vào $(1)$ ta được:
`(3x^2-x+1)(x-1)+x^2-3x^2(x-1)=5/2`
`⇔(x-1)(3x^2-x+1-3x^2)+x^2=5/2`
`⇔(x-1)(-x+1)+x^2=5/2`
`⇔-(x-1)^2+x^2=5/2`
`⇔-x^2+2x-1+x^2=5/2`
`⇔2x=7/2`
`⇔x=7/4`
Vậy `x=7/4`.