Question

Tìm Max A=\(\frac{x-2}{x^3-x^2-x-2}\)

Answer 1

Ta có:

\(A=\frac{x-2}{x^3-x^2-x-2}\) 

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

   \(=\frac{x-2}{\left(x-2\right)\left(x^2+x+1\right)}\)( ĐKXĐ : \(x\ne2\))

\(\Rightarrow A=\frac{1}{x^2+x+1}\)

Lại có: \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\) \(\forall x\ne2\)

\(\Rightarrow A\le\frac{4}{3}\) \(\forall x\ne2\)

Dấu "=" xảy ra khi:

\(x+\frac{1}{2}=0\)\(\Rightarrow x=\frac{-1}{2}\)( thỏa mãn ĐKXĐ )

Vậy \(A_{max}=\frac{4}{3}\Leftrightarrow x=\frac{-1}{2}\)