Đk : \(x\ne2\)
\(A=\dfrac{x-2}{x^3-x^2-x-2}=\dfrac{x-2}{x^3+x^2+x-2x^2-2x-2}\)
\(=\dfrac{x-2}{x\left(x^2+x+1\right)-2\left(x^2+x+1\right)}=\dfrac{x-2}{\left(x-2\right)\left(X^2+x+1\right)}\)
\(=\dfrac{1}{x^2+x+1}\)
\(=\dfrac{1}{\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{3}{4}}\)
\(=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)
Ta có :
\(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)
=> \(A\le\dfrac{4}{3}\)
=> GTLN của \(A=\dfrac{4}{3}\) khi
\(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{1}{2}\)
Vậy với \(x=-\dfrac{1}{2};\) MaxA \(=\dfrac{4}{3}\)
