Đáp án:

c. \(\left[ \begin{array}{l}
x >  - 2\\
x <  - 3
\end{array} \right.\)

Giải thích các bước giải:

\(DK:x \ne \left\{ { - 2; - 3} \right\}\)

\(\begin{array}{l}
a.P = \left( {\dfrac{{x - 3}}{{x + 2}} - \dfrac{{x - 2}}{{x + 3}} + \dfrac{{2 - x}}{{{x^2} + 5x + 6}}} \right):\left( {\dfrac{{1 - x}}{{x - 1}}} \right)\\
 = \left[ {\dfrac{{{x^2} - 9 - {x^2} + 4 + 2 - x}}{{\left( {x + 2} \right)\left( {x + 3} \right)}}} \right]:\left( {\dfrac{{1 - x}}{{ - \left( {1 - x} \right)}}} \right)\\
 = \dfrac{{ - 1 - x}}{{\left( {x + 2} \right)\left( {x + 3} \right)}}.\left( { - 1} \right)\\
 = \dfrac{{x + 1}}{{\left( {x + 2} \right)\left( {x + 3} \right)}}\\
b.P = \dfrac{{x + 1}}{{\left( {x + 2} \right)\left( {x + 3} \right)}} = \dfrac{{x + 2 - 1}}{{\left( {x + 2} \right)\left( {x + 3} \right)}}\\
 = \dfrac{1}{{x + 3}} - \dfrac{1}{{\left( {x + 2} \right)\left( {x + 3} \right)}}\\
P \in Z\\
 \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{{x + 3}} \in Z\\
\dfrac{1}{{\left( {x + 2} \right)\left( {x + 3} \right)}} \in Z
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x + 3 \in U\left( 1 \right)\\
{x^2} + 5x + 6 \in U\left( 1 \right)
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x + 3 = 1\\
x + 3 =  - 1
\end{array} \right.\\
\left[ \begin{array}{l}
{x^2} + 5x + 6 = 1\\
{x^2} + 5x + 6 =  - 1\left( l \right)
\end{array} \right.
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x =  - 2\\
x =  - 4
\end{array} \right.\\
\left[ \begin{array}{l}
x = \dfrac{{ - 5 + \sqrt 5 }}{2}\\
x = \dfrac{{ - 5 - \sqrt 5 }}{2}
\end{array} \right.
\end{array} \right.
\end{array}\)

⇒ Không tồn tại x TMĐK

\(\begin{array}{l}
c.P > 1\\
 \to \dfrac{{x + 1}}{{\left( {x + 2} \right)\left( {x + 3} \right)}} > 1\\
 \to \dfrac{{x + 1 - {x^2} - 5x - 6}}{{\left( {x + 2} \right)\left( {x + 3} \right)}} > 0\\
 \to \dfrac{{ - {x^2} - 4x - 5}}{{\left( {x + 2} \right)\left( {x + 3} \right)}} > 0\\
 \to \left( {x + 2} \right)\left( {x + 3} \right) > 0\left( {do: - {x^2} - 4x - 5 < 0\forall x} \right)\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 2 > 0\\
x + 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 < 0\\
x + 3 < 0
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x >  - 2\\
x <  - 3
\end{array} \right.
\end{array}\)