a) \(M=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right).\dfrac{x-\sqrt{x}}{2\sqrt{x}+1}\left(đk:x\ge0,x\ne1\right)\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}.\dfrac{\sqrt{x}}{2\sqrt{x}+1}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
b) \(M=\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{4}}{\sqrt{4}+1}=\dfrac{2}{2+1}=\dfrac{2}{3}\)
c) \(M=\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{1}{3}\)
\(\Leftrightarrow3\sqrt{x}=\sqrt{x}+1\)
\(\Leftrightarrow2\sqrt{x}=1\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(M=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right)\cdot\dfrac{x-\sqrt{x}}{2\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
b: Thay x=4 vào M, ta được:
\(M=\dfrac{2}{2+1}=\dfrac{2}{3}\)
