21) \(x^2-6x+17=x^2-6x+9+8=\left(x+3\right)^2+8\ge8,\forall x\in R\)
\(\Rightarrow A=\dfrac{1}{x^2-6x+17}\le\dfrac{1}{8},\forall x\in R\)
\(\Rightarrow GTLN\left(A\right)=\dfrac{1}{8}\left(tại.x=-3\right)\)
22) \(\dfrac{x^3-x^2-x-2}{x-2}=x^2+x+1=x^2+x+\dfrac{1}{4}-\dfrac{1}{4}+1\)
\(=\left(x+\dfrac{1}{2}\right)+\dfrac{3}{4}\ge\dfrac{3}{4},\forall x\ne2\)
\(\Rightarrow A=\dfrac{x-2}{x^3-x^2-x-2}\le\dfrac{4}{3},\forall x\ne2\)
\(\Rightarrow GTLN\left(A\right)=\dfrac{4}{3}\left(tại.x=-\dfrac{1}{2}\right)\)
23) \(...A=3-\dfrac{2x}{x^2+1}\)
Ta có : \(\left(x-1\right)^2\ge0\Leftrightarrow x^2+1\ge2x\Leftrightarrow\dfrac{2x}{x^2+1}\le1\)
\(\Rightarrow A=3-\dfrac{2x}{x^2+1}\le3-1=2\)
\(\Rightarrow GTLN\left(A\right)=2\left(tại.x=1\right)\)
