Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
M = \left( {\dfrac{{x + 2}}{{2x - 4}} - \dfrac{{x - 2}}{{2x + 4}} - \dfrac{8}{{4 - {x^2}}}} \right):\dfrac{{4\left( {5x - 3} \right)}}{{7x - 14}}\\
= \left( {\dfrac{{x + 2}}{{2\left( {x - 2} \right)}} - \dfrac{{x - 2}}{{2\left( {x + 2} \right)}} + \dfrac{8}{{\left( {x - 2} \right)\left( {x + 2} \right)}}} \right):\dfrac{{4\left( {5x - 3} \right)}}{{7\left( {x - 2} \right)}}\\
= \dfrac{{{{\left( {x + 2} \right)}^2} - {{\left( {x - 2} \right)}^2} + 16}}{{2.\left( {x - 2} \right)\left( {x + 2} \right)}}:\dfrac{{4\left( {5x - 3} \right)}}{{7\left( {x - 2} \right)}}\\
= \dfrac{{\left( {{x^2} + 4x + 4} \right) - \left( {{x^2} - 4x + 4} \right) + 16}}{{2\left( {x - 2} \right)\left( {x + 2} \right)}}:\dfrac{{4\left( {5x - 3} \right)}}{{7\left( {x - 2} \right)}}\\
= \dfrac{{8x + 16}}{{2.\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{7.\left( {x - 2} \right)}}{{4.\left( {5x - 3} \right)}}\\
= \dfrac{{8.\left( {x + 2} \right)}}{{2\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{7.\left( {x - 2} \right)}}{{4.\left( {5x - 3} \right)}}\\
= \dfrac{7}{{5x - 3}}\\
2,\\
\left\{ \begin{array}{l}
M \in Z\\
2 < M \le 3
\end{array} \right. \Rightarrow M = 3\\
\Rightarrow \dfrac{7}{{5x - 3}} = 3\\
\Leftrightarrow 5x - 3 = \dfrac{7}{3}\\
\Leftrightarrow 5x = \dfrac{{16}}{3}\\
\Leftrightarrow x = \dfrac{{16}}{{15}}\\
3,\\
M > 0 \Leftrightarrow \dfrac{7}{{5x - 3}} > 0 \Leftrightarrow 5x - 3 > 0 \Leftrightarrow x > \dfrac{3}{5}\\
\Rightarrow \left\{ \begin{array}{l}
x > \dfrac{3}{5}\\
x \ne 2
\end{array} \right.
\end{array}\)
