Đáp án:
$\begin{array}{l}
d)\dfrac{{3x - 4}}{{x - 2}} > 1\\
\Rightarrow \dfrac{{3x - 4}}{{x - 2}} - 1 > 0\\
\Rightarrow \dfrac{{3x - 4 - \left( {x - 2} \right)}}{{x - 2}} > 0\\
\Rightarrow \dfrac{{2x - 2}}{{x - 2}} > 0\\
\Rightarrow \dfrac{{x - 1}}{{x - 2}} > 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 1 > 0\\
x - 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 1 < 0\\
x - 2 < 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 1\\
x > 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 1\\
x < 2
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x > 2\\
x < 1
\end{array} \right.\\
Vậy\,x < 1\,hoặc\,x > 2\\
e)\dfrac{{2x - 5}}{{2 - x}} \ge - 1\\
\Rightarrow \dfrac{{2x - 5}}{{2 - x}} + 1 \ge 0\\
\Rightarrow \dfrac{{2x - 5 + 2 - x}}{{2 - x}} \ge 0\\
\Rightarrow \dfrac{{x - 3}}{{2 - x}} \ge 0\\
\Rightarrow \dfrac{{x - 3}}{{x - 2}} \le 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 3 \le 0\\
x - 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 3 \ge 0\\
x - 2 < 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \le 3\\
x > 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x \ge 3\\
x < 2
\end{array} \right.
\end{array} \right.\\
\Rightarrow 2 < x \le 3\\
Vậy\,2 < x \le 3
\end{array}$
