Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\left( {3x + 1} \right)\left( {2 - x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
3x + 1 = 0\\
2 - x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
3x = 1\\
x = 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{3}\\
x = 2
\end{array} \right.\\
b,\\
{x^2} - 36 = 0\\
\Leftrightarrow {x^2} - {6^2} = 0\\
\Leftrightarrow \left( {x - 6} \right)\left( {x + 6} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 6 = 0\\
x + 6 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 6\\
x = - 6
\end{array} \right.\\
c,\\
{\left( {2x + 3} \right)^2} - 4.\left( {{x^2} + 1} \right) = 0\\
\Leftrightarrow {\left( {2x} \right)^2} + 2.2x.3 + {3^2} - \left( {4{x^2} + 4} \right) = 0\\
\Leftrightarrow 4{x^2} + 12x + 9 - 4{x^2} - 4 = 0\\
\Leftrightarrow 12x + 5 = 0\\
\Leftrightarrow 12x = - 5\\
\Leftrightarrow x = - \dfrac{5}{{12}}\\
d,\\
2x - 2 + 3x\left( {x - 1} \right) = 0\\
\Leftrightarrow 2.\left( {x - 1} \right) + 3x\left( {x - 1} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {2 + 3x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
2 + 3x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - \dfrac{2}{3}
\end{array} \right.
\end{array}\)
