a) ĐKXĐ: x≠1; x>0
\(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)^2}=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\sqrt{x}+1\right)=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}+1\right)=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)b) Ta có x=\(6+2\sqrt{5}\Rightarrow A=\dfrac{\left(\sqrt{6+2\sqrt{5}}+1\right)^2}{\sqrt{6+2\sqrt{5}}.\left(\sqrt{6+2\sqrt{5}}-1\right)}=\dfrac{\left(\sqrt{5}+1+1\right)^2}{\left(\sqrt{5}+1\right).\left(\sqrt{5}+1-1\right)}=\dfrac{5+4+4\sqrt{5}}{\left(\sqrt{5}+1\right).\sqrt{5}}=\dfrac{9+4\sqrt{5}}{5+\sqrt{5}}=\dfrac{25+11\sqrt{5}}{20}\)Ta có A<1⇒\(\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}< 1\)
TH1: Nếu \(\sqrt{x}\left(\sqrt{x}-1\right)>0\Rightarrow\left(\sqrt{x}+1\right)^2< \sqrt{x}\left(\sqrt{x}-1\right)\Rightarrow x+2\sqrt{x}+1< x-\sqrt{x}\Rightarrow3\sqrt{x}+1< 0\)
(vô lý)
TH2: Nếu \(\sqrt{x}\left(\sqrt{x}-1\right)< 0\Rightarrow\left(\sqrt{x}+1\right)^2>\sqrt{x}\left(\sqrt{x}-1\right)\Rightarrow3\sqrt{x}+1>0\Rightarrow x>0\)Ta có \(\sqrt{x}\left(\sqrt{x}-1\right)< 0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>0\\\sqrt{x}-1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 0\\\sqrt{x}-1>0\end{matrix}\right.\end{matrix}\right.\)
Cái ngoặc nhọn thứ 2 bị loại
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}>0\\\sqrt{x}-1< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>0\\x< 1\end{matrix}\right.\)(nhận)
Vậy 0<x<1 thì A<1
