Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne 3;x \ne - 5\\
\frac{1}{{\left| {x - 3} \right|}} \le \frac{2}{{x + 5}}\\
\Rightarrow \left[ \begin{array}{l}
\frac{1}{{x - 3}} \le \frac{2}{{x + 5}}\left( {khi:x > 3} \right)\\
\frac{1}{{3 - x}} \le \frac{2}{{x + 5}}\left( {khi:x < 3} \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\frac{{x + 5 - 2\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 5} \right)}} \le 0\left( {x > 3} \right)\\
\frac{{x + 5 - 2\left( {3 - x} \right)}}{{\left( {3 - x} \right)\left( {x + 5} \right)}} \le 0\left( {x < 3} \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\frac{{11 - x}}{{x + 5}} \le 0\left( {x > 3} \right)\\
\frac{{3x - 1}}{{x + 5}} \le 0\left( {x < 3} \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x \ge 11\\
- 5 < x \le \frac{1}{3}
\end{array} \right.\\
b)\frac{3}{{\left| {x + 2} \right|}} \le \frac{1}{{\left| {x - 5} \right|}}\\
\Rightarrow 3\left| {x - 5} \right| \le \left| {x + 2} \right|\\
\Rightarrow \left[ \begin{array}{l}
3\left( {x - 5} \right) \le x + 2\left( {x > 5} \right)\\
3\left( {5 - x} \right) \le x + 2\left( { - 2 < x < 5} \right)\\
3\left( {5 - x} \right) \le - x - 2\left( {x < - 2} \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x \le \frac{{17}}{2}\left( {x > 5} \right)\\
x \ge \frac{{13}}{4}\left( { - 2 < x < 5} \right)\\
x \ge \frac{{17}}{2}\left( {x < - 2} \right)
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
5 < x \le \frac{{17}}{2}\\
\frac{{13}}{4} < x < 5
\end{array} \right.
\end{array}$
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