Question

tim gtln 

\(\frac{x-2}{x^3-x^2-x-2}\)

Answer 1

Ta có :

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

Để \(A\) đạt GTLN \(\Leftrightarrow x^2+x+1\) đạt GTNN

Ta có : \(x^2+x+1=\left(x^2+2\cdot\frac{1}{2}\cdot x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\) có GTNN là 3/4

\(\Rightarrow A\le\frac{1}{\frac{3}{4}}=\frac{4}{3}\) có GTLN là \(\frac{4}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow x=-\frac{1}{2}\)

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