* GTLN:
\(B-3=\dfrac{x^2+x+1}{x^2-x+1}-3=\dfrac{x^2+x+1-3\left(x^2-x+1\right)}{x^2-x+1}\)
\(=\dfrac{x^2+x+1-3x^2+3x-3}{x^2-x+1}=\dfrac{-2x^2+4x-2}{x^2-x+1}\)
\(=\dfrac{-2\left(x^2-2x+1\right)}{x^2-x+1}=\dfrac{-2\left(x-1\right)^2}{x^2-x+1}\le0\)(*)
\(\Rightarrow B-3\le0\Leftrightarrow B\le3\)
Vậy \(B_{max}=3\Leftrightarrow x=1\)
* GTNN:
\(B-\dfrac{1}{3}=\dfrac{x^2+x+1}{x^2-x+1}-\dfrac{1}{3}=\dfrac{x^2+x+1-\dfrac{1}{3}\left(x^2-x+1\right)}{x^2-x+1}\)
\(=\dfrac{x^2+x+1-\dfrac{1}{3}x^2+\dfrac{1}{3}x-\dfrac{1}{3}}{x^2-x+1}=\dfrac{\dfrac{2}{3}x^2+\dfrac{4}{3}x+\dfrac{2}{3}}{x^2-x+1}\)
\(=\dfrac{\dfrac{2}{3}\left(x^2+2x+1\right)}{x^2-x+1}=\dfrac{\dfrac{2}{3}\left(x+1\right)^2}{x^2-x+1}\ge0\)(**)
\(\Rightarrow B-\dfrac{1}{3}\ge0\Leftrightarrow B\ge\dfrac{1}{3}\)
Vậy \(B_{min}=\dfrac{1}{3}\Leftrightarrow x=-1\)
Có (*) và (**) là do \(x^2-x+1=\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
Vậy kết luận chung \(\dfrac{1}{3}\le B\le3\)
