Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \frac{{2x + 6}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\,\,\,\,\,\left( {\left\{ \begin{array}{l}
x \ne - 3\\
x \ne 2
\end{array} \right.} \right)\\
A = 0 \Leftrightarrow 2x + 6 = 0 \Leftrightarrow x = - 3\,\,\,\left( {ko\,\,t/m} \right)\\
A = \frac{{2x + 6}}{{\left( {x + 3} \right)\left( {x - 2} \right)}} = \frac{{2\left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {x - 2} \right)}} = \frac{2}{{x - 2}}\\
B = \frac{{{x^2} - 9}}{{{x^2} - 6x + 9}}\,\,\,\,\left( {x \ne 3} \right)\\
B = 0 \Leftrightarrow {x^2} - 9 = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 3\,\,\,\left( L \right)\\
x = - 3\,\,\left( {t/m} \right)
\end{array} \right.\\
C = \frac{{9{x^2} - 16}}{{3{x^2} - 4x}}\,\,\,\,\,\left( {\left\{ \begin{array}{l}
x \ne 0\\
x \ne \frac{4}{3}
\end{array} \right.} \right)\\
C = 0 \Leftrightarrow 9{x^2} - 16 = 0 \Leftrightarrow \left[ \begin{array}{l}
x = \frac{4}{3}\left( L \right)\\
x = - \frac{4}{3}\left( {t/m} \right)
\end{array} \right.\\
D = \frac{{{x^2} + 4x + 4}}{{2x + 4}}\,\,\,\left( {x \ne - 2} \right)\\
D = 0 \Leftrightarrow {x^2} + 4x + 4 = 0 \Leftrightarrow x = - 2\,\,\left( L \right)\\
D = \frac{{{{\left( {x + 2} \right)}^2}}}{{2\left( {x + 2} \right)}} = \frac{{x + 2}}{2}
\end{array}\)
