\(ĐK:x>0\)
\(M=\left(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\right):\frac{\sqrt{x}}{x+\sqrt{x}}=\left(\frac{\sqrt{x}+1}{x+\sqrt{x}}+\frac{x}{x+\sqrt{x}}\right):\frac{\sqrt{x}}{x+\sqrt{x}}=\frac{x+\sqrt{x}+1}{x+\sqrt{x}}:\frac{\sqrt{x}}{x+\sqrt{x}}=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)
\(x=4\Rightarrow M=\frac{7}{2}\)
\(M=\frac{13}{3}\Leftrightarrow\frac{x+\sqrt{x}+1}{\sqrt{x}}=\frac{13}{3}\Leftrightarrow\frac{x+1}{\sqrt{x}}=\frac{10}{3}\Leftrightarrow3x+3=10\sqrt{x}\Leftrightarrow3x-10\sqrt{x}+3=0\Leftrightarrow x-\frac{10}{3}\sqrt{x}+1=0\Leftrightarrow x-\frac{2.5}{3}\sqrt{x}+\frac{25}{9}=\frac{16}{9}\Leftrightarrow\left(\sqrt{x}-\frac{5}{3}\right)^2=\frac{16}{9}\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-\frac{5}{3}=\frac{4}{3}\\\sqrt{x}-\frac{5}{3}=\frac{-4}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\frac{8}{3}\\\sqrt{x}=\frac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{64}{9}\\x=\frac{1}{9}\end{matrix}\right.\) \(Vậy:x\in\left\{\frac{1}{9};\frac{64}{9}\right\}\)
