\(A=\dfrac{5}{x^2-6x+1}=\dfrac{5}{\left(x^2-6x+9\right)-8}\)
Ta có :
\(\dfrac{5}{\left(x-3\right)^2-8}\le\dfrac{-5}{8}\)vì \(\left(x-3\right)^2-8\ge-8\)
Vậy \(Max_A=\dfrac{-5}{8}\) khi \(x-3=0\Rightarrow x=3\)
\(A=\dfrac{5}{x^2-6x+1}=\dfrac{5}{x^2-6x+9-8}=\dfrac{5}{\left(x-3\right)^2-8}\)
Để A lớn nhất thì \(\left(x-3\right)^2-8\) nhỏ nhất
Ta có: \(\left(x-3\right)^2\ge0\Leftrightarrow\left(x-3\right)^2-8\ge-8\)
\(\Leftrightarrow A=\dfrac{5}{\left(x-3\right)^2-8}\le\dfrac{-5}{8}\)
Dấu " = " khi \(\left(x-3\right)^2=0\Leftrightarrow x=3\)
Vậy \(MAX_A=\dfrac{-5}{8}\) khi x = 3
