Question

Chứng minh rằng:
\(\dfrac{1}{3}\le\dfrac{x^2+x+1}{x^2-x+1}\le3\)
Giải chi tiết giúp mik ạ

Answer 1

\(\dfrac{x^2+x+1}{x^2-x+1}-\dfrac{1}{3}=\dfrac{3x^2+3x+3-x^2+x-1}{3\left(x^2-x+1\right)}\)

\(=\dfrac{2x^2+4x+2}{3\left(x^2-x+1\right)}=\dfrac{2\left(x+1\right)^2}{3\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}}\ge0\)

Do đó: \(\dfrac{1}{3}\le\dfrac{x^2+x+1}{x^2-x+1}\)(1)

\(\dfrac{x^2+x+1}{x^2-x+1}-3=\dfrac{x^2+x+1-3x^2+3x-3}{x^2-x+1}\)

\(=\dfrac{-2x^2+4x-2}{x^2-x+1}=\dfrac{-2\left(x-1\right)^2}{x^2-x+1}\le0\)

Do đó: \(\dfrac{x^2+x+1}{x^2-x+1}\le3\)(2)

Từ (1)và (2) suy ra ĐPCM