Question

Tìm Max của \(B=\dfrac{2x+1}{x^2+2}\)

Answer 1

Trả lời lại

Ta có \(B=\dfrac{2x+1}{x^2+2}\)

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

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

\(\Rightarrow MAX_B=1\Leftrightarrow x=1\)

Answer 2

Ta có B=\(\dfrac{2x+1}{x^2+2}\)

\(\Rightarrow B+\dfrac{1}{2}=\dfrac{2x+1}{x^2+2}+\dfrac{1}{2}\)

\(\Rightarrow B+\dfrac{1}{2}=\dfrac{x^2+4x+4}{x^2+2}=\dfrac{\left(x+2\right)^2}{x^2+2}\ge0\)

\(\Rightarrow MAX_B=-\dfrac{1}{2}\Leftrightarrow x=-2\)