Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
13,\\
\frac{2}{{1 - x}} < 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ne 1} \right)\\
\Leftrightarrow \frac{2}{{1 - x}} - 1 < 0\\
\Leftrightarrow \frac{{2 - \left( {1 - x} \right)}}{{1 - x}} < 0\\
\Leftrightarrow \frac{{x + 1}}{{1 - x}} < 0\\
\Leftrightarrow \frac{{x + 1}}{{x - 1}} > 0\\
\Leftrightarrow \left[ \begin{array}{l}
x > 1\\
x < - 1
\end{array} \right.\\
\Rightarrow S = \left( { - \infty ; - 1} \right) \cup \left( {1; + \infty } \right)\\
14,\\
a,\\
\left| x \right| < 2 \Leftrightarrow - 2 < x < 2\\
b,\\
\left( {x - 1} \right)\left( {x + 2} \right) > 0 \Leftrightarrow \left[ \begin{array}{l}
x > 1\\
x < - 2
\end{array} \right.\\
c,\\
\frac{x}{{1 - x}} + \frac{{1 - x}}{x} < 0\\
\Leftrightarrow \frac{{{x^2} + {{\left( {1 - x} \right)}^2}}}{{\left( {1 - x} \right).x}} < 0\\
{x^2} + {\left( {1 - x} \right)^2} > 0\\
\Rightarrow \left( {1 - x} \right)x < 0 \Leftrightarrow x\left( {x - 1} \right) > 0 \Leftrightarrow \left[ \begin{array}{l}
x > 1\\
x < 0
\end{array} \right.\\
d,\\
\sqrt {x + 3} < x\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge - 3} \right)\\
\Leftrightarrow x + 3 < {x^2}\\
\Leftrightarrow {x^2} - x - 3 > 0\\
\Leftrightarrow \left[ \begin{array}{l}
x > \frac{{1 + \sqrt {13} }}{2}\\
x < \frac{{1 - \sqrt {13} }}{2}
\end{array} \right.\\
15,\\
x + \sqrt {x - 2} \le 2 + \sqrt {x - 2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 2} \right)\\
\Leftrightarrow x \le 2\\
\Rightarrow x = 2\\
16,\\
a,\\
\left( {x + 3} \right)\left( {x + 2} \right) > 0 \Leftrightarrow \left[ \begin{array}{l}
x > - 2\\
x < - 3
\end{array} \right.\\
b,\\
{\left( {x + 3} \right)^2}\left( {x + 2} \right) \le 0 \Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
x + 2 \le 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x \le - 2
\end{array} \right. \Leftrightarrow x \le - 2\\
c,\\
x + \sqrt {1 - {x^2}} \ge 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {1 - {x^2} \ge 0 \Leftrightarrow - 1 \le x \le 1} \right)\\
x = - 3 \Rightarrow 1 - {x^2} < 0\,\,\,\,\,\left( L \right)\\
d,\\
\frac{1}{{1 + x}} + \frac{2}{{3 + 2x}} > 0\\
x = - 3 \Rightarrow \frac{1}{{1 - 3}} + \frac{2}{{3 - 2.3}} > 0 \Leftrightarrow \frac{{ - 7}}{6} > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\left( L \right)\\
17,\\
\frac{{2 - x}}{{2x + 1}} \ge 0\,\,\,\,\,\,\,\,\,\,\,\left( {x \ne - \frac{1}{2}} \right)\\
\Leftrightarrow \frac{{x - 2}}{{2x + 1}} \le 0\\
\Leftrightarrow - \frac{1}{2} < x \le 2
\end{array}\)
