Đáp án:
$\begin{array}{l}
a)\left( {{x^2} - 3x - 5} \right)\left( {x - 2} \right)\\
= {x^3} - 2{x^2} - 3{x^2} + 6x - 5x + 10\\
= {x^3} - 5{x^2} + x + 10\\
b)\left( {\dfrac{1}{3}x - 5} \right).\left( {{x^2} - 3x + 1} \right)\\
= \dfrac{1}{3}{x^3} - {x^2} + \dfrac{1}{3}x - 5{x^2} + 15x - 5\\
= \dfrac{1}{3}{x^3} - 6{x^2} + \dfrac{{46}}{3}x - 5\\
2)\\
a)x\left( {3x - 5} \right) - \dfrac{1}{3}x\left( {4 + 9x} \right) = 5\\
\Rightarrow 3{x^2} - 15x - \dfrac{4}{3}x - 3{x^2} - 5 = 0\\
\Rightarrow \dfrac{{ - 49}}{3}x - 5 = 0\\
\Rightarrow x = \dfrac{{ - 15}}{{49}}\\
b)\left( {1 - 2{x^2}} \right).3x + 2x\left( {3x - 4} \right) = 7 - 4x\\
\Rightarrow 3x - 6{x^3} + 6{x^2} - 8x - 7 + 4x = 0\\
\Rightarrow - 6{x^3} + 6{x^2} - x - 7 = 0\\
\Rightarrow x = - 0,76
\end{array}$
