Giải thích các bước giải:
\(\begin{array}{l}
a,\\
P = \frac{{x + 2}}{{x - 1}}\,\,\,\,\,\,\left( {x \ne 1} \right)\\
\left| {x - 2} \right| = 1 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 1\\
x - 2 = - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 1\,\,\,\,\,\,\left( L \right)
\end{array} \right.\\
x = 3 \Rightarrow P = \frac{{3 + 2}}{{3 - 1}} = \frac{5}{2}\\
b,\\
Q = \frac{{x - 1}}{x} + \frac{{2x + 1}}{{{x^2} + x}}\,\,\,\,\,\,\,\left( {x \ne 0;\,\,\,\,x \ne - 1} \right)\\
\Leftrightarrow Q = \frac{{x - 1}}{x} + \frac{{2x + 1}}{{x\left( {x + 1} \right)}}\\
\Leftrightarrow Q = \frac{{\left( {x - 1} \right)\left( {x + 1} \right) + 2x + 1}}{{x\left( {x + 1} \right)}}\\
\Leftrightarrow Q = \frac{{{x^2} - 1 + 2x + 1}}{{x\left( {x + 1} \right)}}\\
\Leftrightarrow Q = \frac{{x\left( {x + 2} \right)}}{{x\left( {x + 1} \right)}}\\
\Leftrightarrow Q = \frac{{x + 2}}{{x + 1}}\\
c,\\
\left( {x \ne 0;\,\,\,\,x \ne \pm 1} \right)\\
M = P:Q = \frac{{x + 2}}{{x - 1}}:\frac{{x + 2}}{{x + 1}} = \frac{{x + 2}}{{x - 1}}.\frac{{x + 1}}{{x + 2}} = \frac{{x + 1}}{{x - 1}} = \frac{{\left( {x - 1} \right) + 2}}{{x - 1}} = 1 + \frac{2}{{x - 1}}\\
M \in Z \Leftrightarrow \frac{2}{{x - 1}} \in Z \Rightarrow \left( {x - 1} \right) \in \left\{ { \pm 1;\,\,\, \pm 2} \right\}\\
\Rightarrow x \in \left\{ { - 1;0;2;3} \right\}\\
x \ne 0;\,\,\,x \ne \pm 1 \Rightarrow x \in \left\{ {2;3} \right\}
\end{array}\)
