$P=\dfrac{\sqrt{x}-1}{\sqrt{x}}-9\sqrt{x}=1-(9\sqrt{x}+\dfrac{1}{\sqrt{x}})$
Áp dụng BĐT Cô-si ta có:
$9\sqrt{x}+\dfrac{1}{\sqrt{x}}\geq 2\sqrt{9\sqrt{x}.\dfrac{1}{\sqrt{x}}}=6$
$\Rightarrow P=1-(9\sqrt{x}+\dfrac{1}{\sqrt{x}})\leq 1-6=-5$
Dấu '=' xảy ra $\Leftrightarrow x=\dfrac19$
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