Để \(P>\sqrt{x}+1\) thì \(\dfrac{x+5}{\sqrt{x}+2}>\sqrt{x}+1\Leftrightarrow\dfrac{x+5}{\sqrt{x}+2}-\sqrt{x}-1>0\)
<=> \(\dfrac{x+5}{\sqrt{x+2}}-\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}+2}-\dfrac{\sqrt{x}+2}{\sqrt{x}+2}>0\)
<=> \(\dfrac{x+5-x-2\sqrt{x}-\sqrt{x}-2}{\sqrt{x}+2}>0\)
<=> \(\dfrac{-3\sqrt{x}+3}{\sqrt{x}+2}>0\)
Do \(\sqrt{x}+2>0\)
=> \(-3\sqrt{x}+3>0\)
<=> \(-3\sqrt{x}>-3\)
<=> \(\sqrt{x}< 1\)
<=> x<1
Vậy x<1 thì thỏa mãn đk đề bài