Đáp án:
$\begin{array}{l}
a)P = \dfrac{{x + 2}}{{{x^2} - 5x + 6}} - \dfrac{{x + 3}}{{2 - x}} - \dfrac{{x + 2}}{{x - 3}}\\
Q = \dfrac{{2 - x}}{{x + 1}}\left( {x \ne - 1;x \ne 2;x \ne 3} \right)\\
x = 16\left( {tmdk} \right)\\
\Rightarrow Q = \dfrac{{2 - 16}}{{16 + 1}} = \dfrac{{ - 14}}{{17}}\\
b)P = \dfrac{{x + 2}}{{{x^2} - 5x + 6}} - \dfrac{{x + 3}}{{2 - x}} - \dfrac{{x + 2}}{{x - 3}}\\
= \dfrac{{x + 2}}{{\left( {x - 2} \right)\left( {x - 3} \right)}} + \dfrac{{x + 3}}{{x - 2}} - \dfrac{{x + 2}}{{x - 3}}\\
= \dfrac{{x + 2 + \left( {x + 3} \right)\left( {x - 3} \right) - \left( {x + 2} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x - 3} \right)}}\\
= \dfrac{{x + 2 + {x^2} - 9 - {x^2} + 4}}{{\left( {x - 2} \right)\left( {x - 3} \right)}}\\
= \dfrac{{x - 3}}{{\left( {x - 2} \right)\left( {x - 3} \right)}}\\
= \dfrac{1}{{x - 2}}\\
M = P:Q\\
= \dfrac{1}{{x - 2}}:\dfrac{{2 - x}}{{x + 1}}\\
= \dfrac{1}{{x - 2}}.\dfrac{{x + 1}}{{2 - x}}\\
= - \dfrac{{x + 1}}{{{{\left( {x - 2} \right)}^2}}}\\
c)\dfrac{1}{P} = \dfrac{{ - 3}}{2}\\
\Rightarrow - \dfrac{{{{\left( {x - 2} \right)}^2}}}{{x + 1}} = - \dfrac{3}{2}\\
\Rightarrow 2{\left( {x - 2} \right)^2} = 3\left( {x + 1} \right)\\
\Rightarrow 2{x^2} - 8x + 8 = 3x + 3\\
\Rightarrow 2{x^2} - 11x + 5 = 0\\
\Rightarrow \left( {2x - 1} \right)\left( {x - 5} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\left( {tm} \right)\\
x = 5\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = \dfrac{1}{2};x = 5
\end{array}$
