ĐKXĐ: \(x>0\)
a) Ta có: \(M=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{x+\sqrt{x}}\)
\(=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b) Vì x=16 thỏa mãn ĐKXĐ
nên Thay x=16 vào biểu thức \(M=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\), ta được:
\(M=\dfrac{16+\sqrt{16}+1}{\sqrt{16}}=\dfrac{16+4+1}{4}=\dfrac{21}{4}\)
Vậy: Khi x=16 thì \(M=\dfrac{21}{4}\)
c) Để \(M=\dfrac{13}{3}\) thì \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=\dfrac{13}{3}\)
\(\Leftrightarrow3\left(x+\sqrt{x}+1\right)=13\sqrt{x}\)
\(\Leftrightarrow3x+3\sqrt{x}+3-13\sqrt{x}=0\)
\(\Leftrightarrow3x-10\sqrt{x}+3=0\)
\(\Leftrightarrow3x-\sqrt{x}-9\sqrt{x}+3=0\)
\(\Leftrightarrow\sqrt{x}\left(3\sqrt{x}-1\right)-3\left(3\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left(3\sqrt{x}-1\right)\left(\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3\sqrt{x}-1=0\\\sqrt{x}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3\sqrt{x}=1\\\sqrt{x}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{1}{3}\\x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{9}\left(nhận\right)\\x=9\left(nhận\right)\end{matrix}\right.\)
Vậy: Để \(M=\dfrac{13}{3}\) thì \(x\in\left\{\dfrac{1}{9};9\right\}\)