\(\Delta\ge0\Rightarrow-m^2+m+3\ge0\)
\(m^2-m-3\le0\Leftrightarrow\left(m-\dfrac{1-\sqrt{13}}{2}\right)\left(m-\dfrac{1+\sqrt{13}}{2}\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m-\dfrac{1-\sqrt{13}}{2}\ge0\\m-\dfrac{1+\sqrt{13}}{2}\le0\end{matrix}\right.\\\left\{{}\begin{matrix}m-\dfrac{1-\sqrt{13}}{2}\le0\\m-\dfrac{1+\sqrt{13}}{2}\ge0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}m\ge\dfrac{1-\sqrt{13}}{2}\\m\le\dfrac{1+\sqrt{13}}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}m\le\dfrac{1-\sqrt{13}}{2}\\m\ge\dfrac{1+\sqrt{13}}{2}\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\dfrac{1-\sqrt{13}}{2}\le m\le\dfrac{1+\sqrt{13}}{2}\left(1\right)\\\dfrac{1+\sqrt{13}}{2}\le m\le\dfrac{1-\sqrt{13}}{2}\left(2\right)\end{matrix}\right.\)
(2) vô lý => chọn (1)
Vậy .... đc chưa :))))
De phuong trinh co 2 nghiem thi \(\Delta\ge0\)
\(4^2-4\left(m^2-m+1\right)\ge0\)
\(< =>16-4m^2+4m-4\ge0\)
\(< =>4m^2-4m-12\le0\)
\(< =>\left(2m-1\right)^2\le13\)
\(< =>-\sqrt{13}\le2m-1\le\sqrt{13}\)
\(< =>\dfrac{1-\sqrt{13}}{2}\le m\le\dfrac{\sqrt{13}+1}{2}\)
