Đáp án:
$\begin{array}{l}
1)e)\dfrac{2}{{x - 1}} \le \dfrac{5}{{2x - 1}}\\
\Rightarrow \dfrac{{2\left( {2x - 1} \right) - 5\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {2x - 1} \right)}} \le 0\\
\Rightarrow \dfrac{{3 - x}}{{\left( {x - 1} \right)\left( {2x - 1} \right)}} \le 0\\
\Rightarrow \dfrac{{x - 3}}{{\left( {x - 1} \right)\left( {2x - 1} \right)}} \ge 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 3 \ge 0\\
\left( {x - 1} \right)\left( {2x - 1} \right) < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 3 \le 0\\
\left( {x - 1} \right)\left( {2x - 1} \right) > 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 3\\
\dfrac{1}{2} < x < 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 3\\
\left[ \begin{array}{l}
x > 1\\
x < \dfrac{1}{2}
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x < \dfrac{1}{2}\\
1 < x \le 3
\end{array} \right.\\
2)d)\left| {{x^2} - 1} \right| - 2x < 0\\
\Rightarrow \left| {{x^2} - 1} \right| < 2x\\
\Rightarrow \left\{ \begin{array}{l}
2x > 0\\
- 2x < {x^2} - 1 < 2x
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x > 0\\
{x^2} + 2x - 1 > 0\\
{x^2} - 2x - 1 < 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x > 0\\
{\left( {x + 1} \right)^2} > 2\\
{\left( {x - 1} \right)^2} < 2
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x > 0\\
\left[ \begin{array}{l}
x + 1 > \sqrt 2 \\
x + 1 < - \sqrt 2
\end{array} \right.\\
- \sqrt 2 < x - 1 < \sqrt 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x > 0\\
\left[ \begin{array}{l}
x > \sqrt 2 - 1\\
x < - \sqrt 2 - 1
\end{array} \right.\\
1 - \sqrt 2 < x < \sqrt 2 + 1
\end{array} \right.\\
\Rightarrow \sqrt 2 - 1 < x < \sqrt 2 + 1\\
3)d)\left\{ \begin{array}{l}
{\left( {x - 1} \right)^2} \le 0\\
\left( {x + 1} \right)\left( {x - 2} \right) < 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x - 1 = 0\\
- 1 < x < 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 1\\
- 1 < x < 2
\end{array} \right.\\
\Rightarrow x = 1\\
Vậy\,x = 1
\end{array}$
