Question

tìm Min, Max của \(P=\dfrac{x^2+1}{x^2-x+1}\)

Answer 1

+Ta có:\(3P=\dfrac{3x^2+3}{x^2-x+1}=\dfrac{2x^2-2x+2+x^2+2x+1}{x^2-x+1}\)

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

Dấu "=" xảy ra khi \(x=-1\)

Vậy Min \(P=\dfrac{2}{3}\) tại \(x=-1\)

+Ta có: \(P=\dfrac{x^2+1}{x^2-x+1}=\dfrac{x^2+1-2x^2+2x-2+2x^2-2x+2}{x^2-x+1}\)

\(=2-\dfrac{\left(x-1\right)^2}{x^2-x+1}\le2\)

Dấu "=" xảy ra khi x = 1

Vậy Max P = 2 tại x = 1