Ta có:
\(x-5\sqrt{x}+7=x-5\sqrt{x}+\dfrac{5}{4}+\dfrac{23}{4}=\left(\sqrt{x}-\dfrac{5}{2}\right)^2+\dfrac{23}{4}\)
Ta thấy:
\((\sqrt{x}-\dfrac{5}{2})^2\ge0\forall x\)
\(\Leftrightarrow\left(\sqrt{x}-\dfrac{5}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}\)
\(\Leftrightarrow\dfrac{1}{\left(\sqrt{x}-\dfrac{5}{2}\right)^2+\dfrac{23}{4}}\le\dfrac{1}{\dfrac{23}{4}}=\dfrac{4}{23}\)
hay \(P\le\dfrac{4}{23}\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-\dfrac{5}{2}=0\)
\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\)
\(\Leftrightarrow x=\dfrac{25}{4}\)
Vậy Max P = \(\dfrac{4}{23}\) tại \(x=\dfrac{25}{4}\)
