Đáp án:
4) \(\left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ne \pm 2\\
M = \left[ {\dfrac{{x - 2\left( {x + 2} \right) + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}} \right]:\left( {\dfrac{{{x^2} - 4 + 10 - {x^2}}}{{x + 2}}} \right)\\
= \dfrac{{x - 2x - 4 + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}:\dfrac{6}{{x + 2}}\\
= \dfrac{{ - 6}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x + 2}}{6}\\
= - \dfrac{1}{{x - 2}}\\
2)\left| x \right| = \dfrac{1}{2} \to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = - \dfrac{1}{2}
\end{array} \right.\\
Thay:\left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = - \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
M = - \dfrac{1}{{\dfrac{1}{2} - 2}}\\
M = \dfrac{1}{{\dfrac{1}{2} - 2}}
\end{array} \right. \to \left[ \begin{array}{l}
M = \dfrac{2}{3}\\
M = - \dfrac{2}{3}
\end{array} \right.\\
3)M < 0\\
\to - \dfrac{1}{{x - 2}} < 0\\
\to x - 2 > 0\\
\to x > 2\\
4)M \in Z\\
\Leftrightarrow - \dfrac{1}{{x - 2}} \in Z\\
\Leftrightarrow x - 2 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
x - 2 = 1\\
x - 2 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.
\end{array}\)
