Đáp án:
$\begin{array}{l}
a)f\left( x \right).g\left( x \right)\\
= \left( {3{x^2} - x + 1} \right).\left( {x - 1} \right)\\
= 3{x^3} - 3{x^2} - {x^2} + x + x - 1\\
= 3{x^3} - 4{x^2} + 2x - 1\\
b)f\left( x \right).g\left( x \right) + {x^2}\left( {1 - 3g\left( x \right)} \right) = \dfrac{5}{2}\\
\Leftrightarrow 3{x^3} - 4{x^2} + 2x - 1 + {x^2}\left( {1 - 3\left( {x - 1} \right)} \right) = \dfrac{5}{2}\\
\Leftrightarrow 3{x^3} - 4{x^2} + 2x - 1 + {x^2}\left( {1 - 3x + 3} \right) = \dfrac{5}{2}\\
\Leftrightarrow 3{x^3} - 4{x^2} + 2x - 1 + {x^2}\left( {4 - 3x} \right) = \dfrac{5}{2}\\
\Leftrightarrow 3{x^3} - 4{x^2} + 2x - 1 + 4{x^2} - 3{x^3} = \dfrac{5}{2}\\
\Leftrightarrow 2x - 1 = \dfrac{5}{2}\\
\Leftrightarrow 2x = \dfrac{7}{2}\\
\Leftrightarrow x = \dfrac{7}{4}\\
Vậy\,x = \dfrac{7}{4}
\end{array}$
