ĐKXĐ: x>0
Để \(P=\frac{13}{3}\) thì \(\frac{x+\sqrt{x}+1}{\sqrt{x}}=\frac{13}{3}\)
\(\Leftrightarrow3\left(x+\sqrt{x}+1\right)=13\sqrt{x}\)
\(\Leftrightarrow3x+3\sqrt{x}+3-13\sqrt{x}=0\)
\(\Leftrightarrow3x-10\sqrt{x}+3=0\)
\(\Leftrightarrow3x-9\sqrt{x}-\sqrt{x}+3=0\)
\(\Leftrightarrow3\sqrt{x}\left(\sqrt{x}-3\right)-\left(\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right)\left(3\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3=0\\3\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\3\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\\sqrt{x}=\frac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\left(nhận\right)\\x=\frac{1}{9}\left(nhận\right)\end{matrix}\right.\)
Vậy: Khi \(P=\frac{13}{3}\) thì \(x\in\left\{9;\frac{1}{9}\right\}\)
Ta có P = \(\frac{x+\sqrt{x}+1}{\sqrt{x}}\left(x>0\right)\)
Để P = \(\frac{13}{3}\)
\(\Leftrightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}}=\frac{13}{3}\)
\(\Leftrightarrow3\left(x+\sqrt{x}+1\right)=13\sqrt{x}\)
\(\Leftrightarrow3x+3\sqrt{x}+3=13\sqrt{x}\)
\(\Leftrightarrow3x-10\sqrt{x}+3=0\)
\(\Leftrightarrow3\left(\sqrt{x}-3\right)\left(\sqrt{x}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=\frac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=\frac{1}{9}\end{matrix}\right.\left(tm\right)\)
vậy...
