Đáp án:
$\begin{array}{l}
c)Dkxd:x \ne - 3;x \ne 2\\
{x^2} - 9 = 0\\
\Rightarrow x = 3\left( {do:x \ne - 3} \right)\\
\Rightarrow P = \frac{{x - 4}}{{x - 2}} = \frac{{3 - 4}}{{3 - 2}} = - 1\\
d)P = \frac{{ - 3}}{4}\\
\Rightarrow \frac{{x - 4}}{{x - 2}} = \frac{{ - 3}}{4}\\
\Rightarrow 4\left( {x - 4} \right) = - 3\left( {x - 2} \right)\\
\Rightarrow 4x - 16 = - 3x + 6\\
\Rightarrow 7x = 22\\
\Rightarrow x = \frac{{22}}{7}\left( {tmdk} \right)\\
e)P = \frac{{x - 4}}{{x - 2}} = \frac{{x - 2 - 2}}{{x - 2}} = 1 - \frac{2}{{x - 2}} \in Z\\
\Rightarrow \frac{2}{{x - 2}} \in Z\\
\Rightarrow \left( {x - 2} \right) \in \left\{ { - 2; - 1;1;2} \right\}\\
\Rightarrow x \in \left\{ {0;1;3;4} \right\}\\
Do:x \ne 3\\
\Rightarrow x \in \left\{ {0;1;4} \right\}
\end{array}$
