Đáp án:

\(\begin{array}{l}
C7:\\
a) - \dfrac{3}{{a - 2}}\\
b) - \dfrac{3}{4}\\
C9:\\
a)\dfrac{{x + 1}}{{x - 3}}\\
b)\dfrac{5}{3}\\
c)\left[ \begin{array}{l}
x = 7\\
x =  - 1\\
x = 5\\
x = 1\\
x = 4
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
C7:\\
a)DK:a \ne  \pm 2\\
M = \left( {\dfrac{a}{{{a^2} - 4}} + \dfrac{1}{{a + 2}} - \dfrac{2}{{a - 2}}} \right):\left( {1 - \dfrac{a}{{a + 2}}} \right)\\
 = \dfrac{{a + a - 2 - 2\left( {a + 2} \right)}}{{\left( {a - 2} \right)\left( {a + 2} \right)}}:\dfrac{{a + 2 - a}}{{a + 2}}\\
 = \dfrac{{2a - 2 - 2a - 4}}{{\left( {a - 2} \right)\left( {a + 2} \right)}}.\dfrac{{a + 2}}{2}\\
 = \dfrac{{ - 6}}{{\left( {a - 2} \right)\left( {a + 2} \right)}}.\dfrac{{a + 2}}{2}\\
 =  - \dfrac{3}{{a - 2}}\\
b)Thay:a = 6\\
 \to M =  - \dfrac{3}{{6 - 2}} =  - \dfrac{3}{4}\\
C9:\\
a)DK:x \ne \left\{ {2;3} \right\}\\
Q = \dfrac{{2x - 9}}{{{x^2} - 5x + 6}} - \dfrac{{x + 3}}{{x - 2}} - \dfrac{{2x + 1}}{{3 - x}}\\
 = \dfrac{{2x - 9 - \left( {x + 3} \right)\left( {x - 3} \right) + \left( {2x + 1} \right)\left( {x - 2} \right)}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}\\
 = \dfrac{{2x - 9 - {x^2} + 9 + 2{x^2} - 3x - 2}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}\\
 = \dfrac{{{x^2} - x - 2}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}\\
 = \dfrac{{\left( {x + 1} \right)\left( {x - 2} \right)}}{{\left( {x - 3} \right)\left( {x - 2} \right)}}\\
 = \dfrac{{x + 1}}{{x - 3}}\\
b)Thay:x = 9\\
 \to Q = \dfrac{{9 + 1}}{{9 - 3}} = \dfrac{{10}}{6} = \dfrac{5}{3}\\
c)Q = \dfrac{{x + 1}}{{x - 3}} = \dfrac{{x - 3 + 4}}{{x - 3}}\\
 = 1 + \dfrac{4}{{x - 3}}\\
Q \in Z \to \dfrac{4}{{x - 3}} \in Z\\
 \to x - 3 \in U\left( 4 \right)\\
 \to \left[ \begin{array}{l}
x - 3 = 4\\
x - 3 =  - 4\\
x - 3 = 2\\
x - 3 =  - 2\\
x - 3 = 1\\
x - 3 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 7\\
x =  - 1\\
x = 5\\
x = 1\\
x = 4\\
x = 2\left( l \right)
\end{array} \right.
\end{array}\)