$\begin{array}{l}
a)\,\,\,{x^2} - 5 = 0\\
\, \Leftrightarrow \,{x^2} = 5\\
\, \Leftrightarrow x = \pm \sqrt 5 \\
b)\,\,{\left( {2x + 2} \right)^2} - {\left( {3x - 2} \right)^2} = 0\\
\Leftrightarrow \,\left( {2x + 2 + 3x - 2} \right)\left( {2x + 2 - 3x + 2} \right) = 0\\
\Leftrightarrow 5x.\left( {4 - x} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
5x = 0\\
4 - x = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 4
\end{array} \right.\\
c)\,\,5x + 10x = 0\\
\Leftrightarrow 15x = 0\\
\Leftrightarrow x = 10\\
d)\,\,{x^2} + 4x + 3 = 0\,\\
\Leftrightarrow \,\left( {{x^2} + 3x} \right) + \left( {x + 3} \right) = 0\\
\Leftrightarrow x\left( {x + 3} \right) + \left( {x + 3} \right) = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {x + 1} \right)\, = 0\\
\Rightarrow \left[ \begin{array}{l}
x + 3 = 0\\
x + 1 = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = - 3\\
x = - 1
\end{array} \right.
\end{array}$
Hai câu còn lại em tự suy nghĩ và làm tương tự nhé.
