\(A=\dfrac{2}{2x^2+10x-1}=\dfrac{2}{2\left(x^2+5x+\dfrac{25}{4}\right)-\dfrac{27}{2}}=\dfrac{2}{2\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{2}}\)Vì \(2\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{2}\ge\dfrac{-27}{2}\)
\(\Rightarrow\dfrac{2}{2\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{2}}\le\dfrac{2}{\dfrac{-27}{2}}=\dfrac{-4}{27}\)
Vậy \(Max_A=\dfrac{-4}{27}\) khi \(x+\dfrac{5}{2}=0\Rightarrow x=\dfrac{-5}{2}\)
Mẫu = \(2x^2+10x-1=2\left(x^2+2.\dfrac{5}{2}x+\dfrac{25}{4}\right)-\dfrac{27}{2}\)
\(=2\left(x+\dfrac{5}{2}\right)^2-\dfrac{27}{2}\ge-\dfrac{27}{2}\)
\(=>MIN_{Mẫu}=-\dfrac{27}{2}\)
A lớn nhất khi mẫu nhỏ nhất : \(A=\dfrac{2}{-\dfrac{27}{2}}=-\dfrac{4}{27}\)
\(=>MAX_A=-\dfrac{4}{27}\Leftrightarrow x=-\dfrac{5}{2}\)
