Gọi \(A=\frac{x^2+x+1}{x^2-x+1}\)ta có :
\(A=\frac{\frac{1}{3}x^2-\frac{1}{3}x+\frac{1}{3}+\frac{2}{3}x^2-\frac{4}{3}x+\frac{2}{3}}{x^2-x+1}=\frac{\frac{1}{3}\left(x^2-x+1\right)+\frac{2}{3}\left(x^2-2x+1\right)}{x^2-x+1}\)
\(=\frac{1}{3}+\frac{\frac{2}{3}\left(x-1\right)^2}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\ge\frac{1}{3}\forall x\) có GTNN là \(\frac{1}{3}\) tại \(x=1\)
Sr nhìn lộn
\(A=\frac{x^2+x+1}{x^2-x+1}=\frac{\frac{1}{3}x^2-\frac{1}{3}x+\frac{1}{3}+\frac{2}{3}x^2+\frac{4}{3}x+\frac{2}{3}}{x^2-x+1}=\frac{\frac{1}{3}\left(x^2-x+1\right)+\frac{2}{3}\left(x^2+2x+1\right)}{x^2-x+1}\)
\(=\frac{1}{3}+\frac{\frac{2}{3}\left(x+1\right)^2}{x^2-x+1}\ge\frac{1}{3}\) có gtnn là 1/3 tại x = - 1
