\[\begin{array}{l}
a)\,\,f\left( x \right) = 3{x^2} - 2x + 1\\
co:\,\,\Delta ' = 1 - 3.1 = - 2 < 0\\
Ma\,\,a = 3 > 0\\
\Rightarrow f\left( x \right) > 0\,\,\forall x.\\
b)\,\,f\left( x \right) = 2{x^2} + 5x + 2\\
co\,\,\Delta = {5^2} - 4.2.2 = 9 > 0\\
\Rightarrow da\,\,thuc\,\,f\left( x \right)\,\,\,co\,\,2\,\,nghiem\,\,\,phan\,\,biet:\,\,\left[ \begin{array}{l}
{x_1} = \frac{{ - 5 + \sqrt 9 }}{{2.2}} = - \frac{1}{2}\\
{x_2} = \frac{{ - 5 - \sqrt 9 }}{{2.2}} = - 2
\end{array} \right..\\
\Rightarrow \left[ \begin{array}{l}
f\left( x \right) > 0\,\,\,khi\,\,x \in \left( { - \infty ;\,\, - 2} \right) \cup \left( { - \frac{1}{2}; + \infty } \right)\\
f\left( x \right) < 0\,\,khi\,\,x \in \left( { - 2;\,\, - \frac{1}{2}} \right)
\end{array} \right..
\end{array}\]
Em làm tương tự ở các ý sau nhé e.
