Đáp án:
a) \(\left\{ \begin{array}{l}
x \le \dfrac{{15}}{8}\\
x \ge \dfrac{9}{8}
\end{array} \right.\)
b) \(\left\{ \begin{array}{l}
x \le 2\\
x \ge - 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)4{x^2} - 12x \le - \dfrac{{135}}{{16}}\\
\to 4{x^2} - 12x + 9 \le \dfrac{9}{{16}}\\
\to {\left( {2x - 3} \right)^2} \le \dfrac{9}{{16}}\\
\to \left\{ \begin{array}{l}
2x - 3 \le \dfrac{3}{4}\\
2x - 3 \ge - \dfrac{3}{4}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \le \dfrac{{15}}{8}\\
x \ge \dfrac{9}{8}
\end{array} \right.\\
b){x^2} - x \le 2\\
\to {x^2} - x - 2 \le 0\\
\to \left( {x - 2} \right)\left( {x + 1} \right) \le 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 2 \ge 0\\
x + 1 \le 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 2 \le 0\\
x + 1 \ge 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 2\\
x \le - 1
\end{array} \right.\left( l \right)\\
\left\{ \begin{array}{l}
x \le 2\\
x \ge - 1
\end{array} \right.
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \le 2\\
x \ge - 1
\end{array} \right.
\end{array}\)
