$ĐKXĐ: x\ge0$
$P=\dfrac{1}{x-5\sqrt{x}+7}$
$P=\dfrac{1}{(\sqrt{x})^2-2.\sqrt{x}.\dfrac{5}{2}+(\dfrac{5}{2})^2+\dfrac{3}{4}}$
$P=\dfrac{1}{(\sqrt{x}-\dfrac{5}{2})^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}$
Dấu $"="$ xảy ra
$\Leftrightarrow\sqrt{x}-\dfrac{5}{2}=0$
$\Leftrightarrow\sqrt{x}=\dfrac{5}{2}$
$\Leftrightarrow x=\dfrac{25}{4}(t/m)$
Vậy $MaxP=\dfrac{4}{3}$ khi $x=\dfrac{25}{4}$
