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a, $3x^2+2x-8\le 0$
$3x^2+2x-8=0\Leftrightarrow\left[\begin{array}{1}x=\dfrac{4}{3}\\x=-2\end{array}\right.$
\begin{array}{|c|cc|}\hline x&-\infty&&-2&&\dfrac{4}{3}&&+\infty\\\hline f(x)&&+&0&-&0&+&\\\hline\end{array}
$f(x)\le 0\Rightarrow -2\le x\le \dfrac{4}{3}$
Vậy $-2\le x\le\dfrac{4}{3}$
b, $2x^2+13x+20>0$
$2x^2+13x+20=0\Leftrightarrow\left[\begin{array}{1}x=-\dfrac{5}{2}\\x=-4\end{array}\right.$
\begin{array}{|c|cc|}\hline x&-\infty&&-4&&-\dfrac{5}{2}&&+\infty\\\hline f(x)&&+&0&-&0&+&\\\hline\end{array}
$f(x)>0\Rightarrow x\in (-\infty;-4)∪\left(-\dfrac{5}{2};+\infty\right)$
Vậy $x\in (-\infty;-4)∪\left(-\dfrac{5}{2};+\infty\right)$
c, $3x^2+x+1\le 0$
$3x^2+x+1=3\left(x+\dfrac{1}{6}\right)^2+\dfrac{11}{12}>0\forall x\in\mathbb R$
$\Rightarrow 3x^2+x+1\le 0$ vô nghiệm
Vậy $x\in\varnothing$
d, $-2x^2-3x+1\le 0$
$-2x^2-3x+1=0\Leftrightarrow x=\dfrac{-3\pm\sqrt{17}}{4}$
\begin{array}{|c|cc|}\hline x&-\infty&&\dfrac{-3-\sqrt{17}}{4}&&\dfrac{-3+\sqrt{17}}{4}&&+\infty\\\hline f(x)&&-&0&+&0&-&\\\hline\end{array}
$f(x)\le 0\Rightarrow x\in\left(-\infty;\dfrac{-3-\sqrt{17}}{4}\right]∪\left[\dfrac{-3+\sqrt{17}}{4};+\infty\right)$
Vậy $x\in\left(-\infty;\dfrac{-3-\sqrt{17}}{4}\right]∪\left[\dfrac{-3+\sqrt{17}}{4};+\infty\right)$
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