Đáp án:
$\begin{array}{l}
1)10{x^3} - {x^2} - 3x = 0\\
\Leftrightarrow x\left( {10{x^2} - x - 3} \right) = 0\\
\Leftrightarrow x\left( {10{x^2} + 5x - 6x - 3} \right) = 0\\
\Leftrightarrow x.\left( {2x + 1} \right)\left( {5x - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
2x + 1 = 0\\
5x - 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - \dfrac{1}{2}\\
x = \dfrac{3}{5}
\end{array} \right.\\
Vay\,x = - \dfrac{1}{2};x = 0;x = \dfrac{3}{5}\\
2)\\
A = \dfrac{{{x^2} - 25}}{{{x^2} + 10x + 25}}\\
= \dfrac{{\left( {x - 5} \right)\left( {x + 5} \right)}}{{{{\left( {x + 5} \right)}^2}}} = \dfrac{{x - 5}}{{x + 5}}\\
B = \dfrac{{{x^3} - 27}}{{{x^3} + 3{x^2} + 9x}}\\
= \dfrac{{\left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right)}}{{x\left( {{x^2} + 3x + 9} \right)}}\\
= \dfrac{{x - 3}}{x}\\
C = \dfrac{{3{x^2} - 12}}{{6{x^2} + 24x + 24}}\\
= \dfrac{{3\left( {{x^2} - 4} \right)}}{{6\left( {{x^2} + 4x + 4} \right)}}\\
= \dfrac{{3\left( {x - 2} \right)\left( {x + 2} \right)}}{{6.{{\left( {x + 2} \right)}^2}}}\\
= \dfrac{{x - 2}}{{2\left( {x + 2} \right)}}\\
= \dfrac{{x - 2}}{{2x + 4}}
\end{array}$
