Đáp án:
$\begin{array}{l}
c)5.\left( {x - 3} \right) + 4\left( {x + 2} \right) = 12\\
\Leftrightarrow 5x - 15 + 4x + 8 = 12\\
\Leftrightarrow 9x = 19\\
\Leftrightarrow x = \dfrac{{19}}{9}\\
Vậy\,x = \dfrac{{19}}{9}\\
d){\left( {x - 1} \right)^2} - {\left( {x + 1} \right)^2} = 4\\
\Leftrightarrow \left( {x - 1 + x + 1} \right)\left( {x - 1 - x - 1} \right) = 4\\
\Leftrightarrow 2x.\left( { - 2} \right) = 4\\
\Leftrightarrow x = - 1\\
Vậy\,x = - 1\\
e)5x\left( {x - 3} \right) - x\left( 5 \right).\left( { - 1} \right) = 4\\
\Leftrightarrow 5{x^2} - 15x + 5x - 4 = 0\\
\Leftrightarrow 5{x^2} - 10x - 4 = 0\\
\Leftrightarrow x = \dfrac{{5 \pm 3\sqrt 5 }}{5}\\
Vậy\,x = \dfrac{{5 \pm 3\sqrt 5 }}{5}
\end{array}$
