Đáp án:
a. \(M = \dfrac{7}{{5x - 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)M = \left[ {\dfrac{{{{\left( {x + 2} \right)}^2} - {{\left( {x - 2} \right)}^2} + 8.2}}{{2\left( {x - 2} \right)\left( {x + 2} \right)}}} \right].\dfrac{{7\left( {x - 2} \right)}}{{4\left( {5x - 3} \right)}}\\
= \left[ {\dfrac{{{x^2} + 4x + 4 - {x^2} + 4x - 4 + 16}}{{2\left( {x - 2} \right)\left( {x + 2} \right)}}} \right].\dfrac{{7\left( {x - 2} \right)}}{{4\left( {5x - 3} \right)}}\\
= \dfrac{{8\left( {x + 2} \right)}}{{2\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{7\left( {x - 2} \right)}}{{4\left( {5x - 3} \right)}}\\
= \dfrac{7}{{5x - 3}}\\
2)\left\{ \begin{array}{l}
2 < \dfrac{7}{{5x - 3}}\\
\dfrac{7}{{5x - 3}} \le 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{7 - 10x + 6}}{{5x - 3}} > 0\\
\dfrac{{7 - 15x + 9}}{{5x - 3}} \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{13 - 10x}}{{5x - 3}} > 0\\
\dfrac{{16 - 15x}}{{5x - 3}} \le 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
5x - 3 > 0\\
13 - 10x > 0\\
16 - 15x \le 0
\end{array} \right.\\
\left\{ \begin{array}{l}
5x - 3 < 0\\
13 - 10x < 0\\
16 - 15x \ge 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > \dfrac{3}{5}\\
x < \dfrac{{13}}{{10}}\\
x \ge \dfrac{{16}}{{15}}
\end{array} \right.\\
\left\{ \begin{array}{l}
x < \dfrac{3}{5}\\
x > \dfrac{{13}}{{10}}\\
x \le \dfrac{{16}}{{15}}
\end{array} \right.\left( l \right)
\end{array} \right.\\
KL:\dfrac{{16}}{{15}} \le x < \dfrac{{13}}{{10}}\\
3)M > 0\\
\to \dfrac{7}{{5x - 3}} > 0\\
\to 5x - 3 > 0\\
\to x > \dfrac{3}{5};x \ne 2
\end{array}\)
