Lời giải:
$5x^2-8x+3\geq 0$
$\Leftrightarrow 5x(x-1)-3(x-1)\geq 0$
$\Leftrightarrow (x-1)(5x-3)\geq 0$
\(\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x-1\geq 0\\ 5x-3\geq 0\end{matrix}\right.\\ \left\{\begin{matrix} x-1\leq 0\\ 5x-3\leq 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x\geq 1\\ x\geq \frac{3}{5}\end{matrix}\right.\\ \left\{\begin{matrix} x\leq 1\\ x\leq \frac{3}{5}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x\geq 1\\ x\leq \frac{3}{5}\end{matrix}\right.\)
b.
$3x^2-x-3\leq 0$
$\Leftrightarrow (x-\frac{1+\sqrt{37}}{6})(x-\frac{1-\sqrt{37}}{6})\leq 0$
$\Leftrightarrow$ \(\left\{\begin{matrix} x-\frac{1+\sqrt{37}}{6}\ge 0\\ x-\frac{1-\sqrt{37}}{6}\leq 0\end{matrix}\right.\) hoặc \(\left\{\begin{matrix} x-\frac{1+\sqrt{37}}{6}\leq 0\\ x-\frac{1-\sqrt{37}}{6}\geq 0\end{matrix}\right.\)
\(\Leftrightarrow \frac{1+\sqrt{37}}{6}\leq x\leq \frac{1-\sqrt{37}}{6}\) (vô lý) hoặc \(\frac{1+\sqrt{37}}{6}\geq x\geq \frac{1-\sqrt{37}}{6}\)
Vậy $\frac{1+\sqrt{37}}{6}\geq x\geq \frac{1-\sqrt{37}}{6}$
c.
$x^2\geq 4$
$\Leftrightarrow x^2-4\geq 0$
$\Leftrightarrow (x-2)(x+2)\geq 0$
$\Leftrightarrow x\geq 2$ hoặc $x\leq -2$
