Vì x\(\ge0,x\ne1\) nên \(\sqrt{x}+1\ge1\), do đó, để P>0 thì\(-x+6\sqrt{x}+9>0\Leftrightarrow-\left(x-6\sqrt{x}+9\right)+18>0\Leftrightarrow-\left(\sqrt{x}-3\right)^2>-18\Leftrightarrow\left(\sqrt{x}-3\right)^2< 18\Leftrightarrow\sqrt{\left(\sqrt{x}-3\right)^2}< \sqrt{18}\Leftrightarrow\left|\sqrt{x}-3\right|< 3\sqrt{2}\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3< 3\sqrt{2}\\-\left(\sqrt{x}-3\right)< 3\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}< 3\sqrt{2}+3\\\sqrt{x}>3-3\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< \left(3\sqrt{2}+3\right)^2\\x>\left(3-3\sqrt{2}\right)^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 18+9+18\sqrt{2}\\x>9+18-18\sqrt{2}\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x< 27+18\sqrt{2}\\x>27-18\sqrt{2}\end{matrix}\right.\Leftrightarrow27-18\sqrt{2}< x< 27+18\sqrt{2}\)
Ơ, sao ko tải đc, đáp án cuối cùng là \(27-18\sqrt{2}< x< 27+18\sqrt{2}\)
