xét dấu các tam thức bậc hai sau: a,f(x)=3x^2 -2x +1 b,f(x)=2x^2+5x+2 c,f(x)=-3x^2 +5x+1 d,f(x)=-4x^2+3x+1 e,f(x)=3x^2+x-5

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