\(M=\frac{x^2+1}{x^2-x+1}\)\(\Rightarrow M\left(x^2-x+1\right)=x^2+1\)
\(\Leftrightarrow Mx^2-Mx+M-x^2-1=0\)
\(\Leftrightarrow x^2\left(M-1\right)-Mx-M+1=0\)
\(\Delta=\left(-M\right)^2-4\left(M-1\right)^2\ge0\)
\(\Leftrightarrow M^2-4\left(M^2-2M+1\right)\ge0\)
\(\Leftrightarrow-3M^2+8M-4\ge0\)
\(\Leftrightarrow3M^2-8M+4\le0\)
\(\Leftrightarrow\left(3M-2\right)\left(M-2\right)\le0\)
\(\Leftrightarrow\frac{2}{3}\le M\le2\)
Vậy \(Min_M=\frac{2}{3}\Leftrightarrow x=\frac{-b}{2a}=\frac{M}{2\left(M-1\right)}=\frac{\frac{2}{3}}{2\left(\frac{2}{3}-1\right)}=-1\)
\(Max_M=2\Leftrightarrow x=\frac{-b}{2a}=\frac{M}{2\left(M-1\right)}=\frac{2}{2\left(2-1\right)}=1\)