Đáp án:

$\begin{array}{l}
a){x^2} - 3x + 5 \ge 0\\
 \Leftrightarrow {x^2} - 2.x.\dfrac{3}{2} + \dfrac{9}{4} + \dfrac{{11}}{4} \ge 0\\
 \Leftrightarrow {\left( {x - \dfrac{3}{2}} \right)^2} + \dfrac{{11}}{4} \ge 0\left( {tmdk} \right)\\
Vậy\,x \in R\\
b){x^2} + 5x - 3 \le 0\\
 \Leftrightarrow {x^2} + 2.x.\dfrac{5}{2} + \dfrac{{25}}{4} - \dfrac{{13}}{4} \le 0\\
 \Leftrightarrow {\left( {x + \dfrac{5}{2}} \right)^2} \le \dfrac{{13}}{4}\\
 \Leftrightarrow \dfrac{{ - \sqrt {13}  - 5}}{2} \le x \le \dfrac{{\sqrt {13}  - 5}}{2}\\
Vậy\,\dfrac{{ - \sqrt {13}  - 5}}{2} \le x \le \dfrac{{\sqrt {13}  - 5}}{2}\\
c) - {x^2} + 7x + 3 \ge 0\\
 \Leftrightarrow {x^2} - 7x - 3 \le 0\\
 \Leftrightarrow {x^2} - 2.x.\dfrac{7}{2} + \dfrac{{49}}{4} - \dfrac{{61}}{4} \le 0\\
 \Leftrightarrow {\left( {x - \dfrac{7}{2}} \right)^2} \le \dfrac{{61}}{4}\\
 \Leftrightarrow \dfrac{{7 - \sqrt {61} }}{2} \le x \le \dfrac{{7 + \sqrt {61} }}{2}\\
Vậy\,\dfrac{{7 - \sqrt {61} }}{2} \le x \le \dfrac{{7 + \sqrt {61} }}{2}\\
d)4{x^2} - 5x - 1 \ge 0\\
 \Leftrightarrow 4{x^2} - 2.2x.\dfrac{5}{4} + \dfrac{{25}}{{16}} - \dfrac{{41}}{{16}} \ge 0\\
 \Leftrightarrow {\left( {2x - \dfrac{5}{4}} \right)^2} \ge \dfrac{{41}}{{16}}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{5}{4} \ge \dfrac{{\sqrt {41} }}{4}\\
2x - \dfrac{5}{4} \le \dfrac{{ - \sqrt {41} }}{4}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x \ge \dfrac{{5 + \sqrt {41} }}{8}\\
x \le \dfrac{{5 - \sqrt {41} }}{8}
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x \ge \dfrac{{5 + \sqrt {41} }}{8}\\
x \le \dfrac{{5 - \sqrt {41} }}{8}
\end{array} \right.
\end{array}$