Đáp án:
a) \(x \ge - \dfrac{1}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)36{x^2} - 12x + 1 \le 36{x^2} + 60x + 25\\
\to 72x \ge - 24\\
\to x \ge - \dfrac{1}{3}\\
b)\left\{ \begin{array}{l}
\left( {x + 3} \right)\left( {6x + 1} \right) \le 3x + \dfrac{1}{2}\\
\dfrac{{2 - x}}{4} \ge 2x + \dfrac{{x - 5}}{3}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
6{x^2} + 19x + 3 \le 3x + \dfrac{1}{2}\\
3\left( {2 - x} \right) \ge 12.2x + 4\left( {x - 5} \right)
\end{array} \right.\\
\to \left\{ \begin{array}{l}
6{x^2} + 16x + \dfrac{5}{2} \le 0\\
6 - 3x \ge 24x + 4x - 20
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {6x + 1} \right)\left( {2x + 5} \right) \le 0\\
31x \le 26
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \le \dfrac{{26}}{{31}}\\
- \dfrac{5}{2} \le x \le - \dfrac{1}{6}
\end{array} \right.\\
\to - \dfrac{5}{2} \le x \le - \dfrac{1}{6}
\end{array}\)
