\(\dfrac{2}{x^2+x+1}=\dfrac{2}{x^2+\dfrac{1}{2}x+\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}}\)
\(=\dfrac{2}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(x+\dfrac{1}{2}\right)^2\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(\Rightarrow\dfrac{2}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{8}{3}\)
với mọi giá trị của \(x\in R\)
Để \(\dfrac{2}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}=\dfrac{8}{3}\) thì \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2=0\Rightarrow x=-\dfrac{1}{2}\)
Chúc bạn học tốt!!!
