ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có :
\(Q=\left(\frac{1}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right):\frac{\sqrt{x}-1}{x+2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}+1}-\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right).\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}\)
Vậy ....
~~~ Mình nghĩ khác chị ở dưới và cũng chẳng biết đúng hay sai nữa ~~~
a, \(Đkxđ:\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(Q=\left(\frac{1}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right):\frac{\sqrt{x}-1}{x+2\sqrt{x+1}}\)
\(=\left[\frac{1}{\sqrt{x}+1}-\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right].\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\frac{\sqrt{x}}{\sqrt{x}}+\frac{1}{\sqrt{x}}\)
\(=1+\frac{1}{\sqrt{x}}\)
b, \(Vì:\sqrt{x}>0\forall x>0;x\ne1\)
\(\Rightarrow\frac{1}{\sqrt{x}}>0\forall x\)
\(\Rightarrow\frac{1}{\sqrt{x}}+1>1\forall x\)
\(\Rightarrow Q>1\)
~~~ P/s: Chỉ là nghĩ thôi ~~~~