Sai thông cảm ặ
\(-3x^2+3x+1=-3\left(x^2-x-\frac{1}{3}\right)=-3\left(x^2-2x.\frac{1}{2}+\frac{1}{4}-\frac{7}{12}\right)=-3[\left(x-\frac{1}{2}\right)^2-\frac{7}{12}]\)
Mà để \(-3[\left(x-\frac{1}{2}\right)^2-\frac{7}{12}\)là số dương \(\Leftrightarrow-3[\left(x-\frac{1}{2}\right)^2-\frac{7}{12}]>0\)
Mà \(\left(-3\right)< 0\Rightarrow\left(x-\frac{1}{2}\right)^2-\frac{7}{12}< 0\Rightarrow\left(x-\frac{1}{2}\right)^2< \frac{7}{12}\Rightarrow\left(x-\frac{1}{2}\right)^2< \left(\frac{\sqrt{7}}{2\sqrt{3}}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}>0-\frac{\sqrt{7}}{2\sqrt{3}}\\x-\frac{1}{2}< \frac{\sqrt{7}}{2\sqrt{3}}\end{cases}}\)
\(\Rightarrow\frac{-\sqrt{7}+\sqrt{3}}{2\sqrt{3}}< x< \frac{\sqrt{7}+\sqrt{3}}{2\sqrt{3}}\)thì \(-3x^2+3x+1>0\)