Đáp án:
$\begin{array}{l}
g)6{x^2} - \left( {2x + 5} \right)\left( {3x - 2} \right) = - 12\\
\Leftrightarrow 6{x^2} - \left( {6{x^2} - 4x + 15x - 10} \right) = - 12\\
\Leftrightarrow 6{x^2} - 6{x^2} - 11x + 10 = - 12\\
\Leftrightarrow - 11x = - 22\\
\Leftrightarrow x = 2\\
Vậy\,x = 2\\
h){\left( {2x + 3} \right)^2} - {\left( {x - 1} \right)^2} = 0\\
\Leftrightarrow \left( {2x + 3 - x + 1} \right)\left( {2x + 3 + x - 1} \right) = 0\\
\Leftrightarrow \left( {x + 4} \right)\left( {3x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 4\\
x = - \dfrac{2}{3}
\end{array} \right.\\
Vậy\,x = - 4;x = - \dfrac{2}{3}\\
i)5x\left( {x - 1} \right) - \left( {x + 2} \right)\left( {5x - 7} \right) = 6\\
\Leftrightarrow 5{x^2} - 5x - \left( {5{x^2} - 7x + 10x - 14} \right) = 6\\
\Leftrightarrow 5{x^2} - 5x - 5{x^2} - 3x + 14 = 6\\
\Leftrightarrow - 8x = - 8\\
\Leftrightarrow x = 1\\
Vậy\,x = 1\\
k){\left( {x + 2} \right)^2} - \left( {{x^2} - 4} \right) = 0\\
\Leftrightarrow {x^2} + 4x + 4 - {x^2} + 4 = 0\\
\Leftrightarrow 4x = - 8\\
\Leftrightarrow x = - 2\\
Vậy\,x = - 2\\
l)x\left( {x + 5} \right) - \left( {x - 2} \right)\left( {x + 3} \right) = 0\\
\Leftrightarrow {x^2} + 5x - \left( {{x^2} + 3x - 2x - 6} \right) = 0\\
\Leftrightarrow {x^2} + 5x - {x^2} - x + 6 = 0\\
\Leftrightarrow 4x = - 6\\
\Leftrightarrow x = - \dfrac{3}{2}\\
Vậy\,x = - \dfrac{3}{2}
\end{array}$
