\(C=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)=\dfrac{x+\sqrt{x}+1}{x+1}:\dfrac{x-2\sqrt{x}+1}{\left(x+1\right)\left(\sqrt{x}-1\right)}=\dfrac{x+\sqrt{x}+1}{x+1}.\dfrac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}\left(x\ne1;x\ge0\right)\)
\(a.C>3\Leftrightarrow\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}-3>0\)
\(\Leftrightarrow\dfrac{x-2\sqrt{x}+4}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}-1\right)^2+3}{\sqrt{x}-1}>0\)
\(\Leftrightarrow x>1\)
\(b.x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\left(TM\right)\)
\(\Rightarrow\sqrt{x}=\sqrt{3}+1\)
Khi đó : \(C=\dfrac{4+2\sqrt{3}+\sqrt{3}+1+1}{\sqrt{3}+1-1}=\dfrac{6+3\sqrt{3}}{\sqrt{3}}=\dfrac{3\left(2+\sqrt{3}\right)}{\sqrt{3}}=\sqrt{3}\left(2+\sqrt{3}\right)\)
KL.........
