Đáp án:
$\begin{array}{l}
a)\left| {x + 1} \right| > x + 3\\
+ Khi:x + 3 < 0\left( {luon\,dung} \right)\\
\Rightarrow x < - 3\\
+ Khi:x + 3 \ge 0 \Rightarrow x \ge - 3\\
\Rightarrow {\left( {x + 1} \right)^2} > {\left( {x + 3} \right)^2}\\
\Rightarrow {x^2} + 2x + 1 > {x^2} + 6x + 9\\
\Rightarrow 4x < - 8\\
\Rightarrow x < - 2\left( {ktm} \right)\\
\Rightarrow - 3 \le x < - 2\\
Vay\,x < - 2\\
b)\left| {2x - 1} \right| < x + 2\\
Dk:x + 2 > 0\\
\Rightarrow x > - 2\\
\Rightarrow {\left( {2x - 1} \right)^2} < {\left( {x + 2} \right)^2}\\
\Rightarrow 4{x^2} - 4x + 1 < {x^2} + 4x + 4\\
\Rightarrow 3{x^2} < 3\\
\Rightarrow {x^2} < 1\\
\Rightarrow - 1 < x < 1\\
Vay\, - 1 < x < 1\\
c)\dfrac{3}{{2 - x}} \le 1\\
\Rightarrow \dfrac{{3 - 2 + x}}{{2 - x}} \le 0\\
\Rightarrow \dfrac{{x + 1}}{{2 - x}} \le 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 \le 0\\
2 - x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 \ge 0\\
2 - x < 0
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \le - 1\\
x < 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x \ge - 1\\
x > 2
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x \le - 1\\
x > 2
\end{array} \right.
\end{array}$
