Đáp án:
$\begin{array}{l}
1)\left( {{x^2} - 3x + 2} \right)\left( {\dfrac{1}{3}x - 4} \right)\\
= \dfrac{1}{3}{x^3} - 4{x^2} - {x^2} + 12x + \dfrac{2}{3}x - 8\\
= \dfrac{1}{3}{x^3} - 5{x^2} + \dfrac{{38}}{3}x - 8\\
2)\\
2x\left( {x - 5} \right) - x.\left( {3 + 2x} \right) = 26\\
\Leftrightarrow 2{x^2} - 10x - 3x - 2{x^2} = 26\\
\Leftrightarrow - 13x = 26\\
\Leftrightarrow x = - 2\\
Vậy\,x = - 2
\end{array}$
