a: \(B=\dfrac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\left(\sqrt{x}+1\right)=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b: \(x=4-2\sqrt{3}+2-\sqrt{3}=6-3\sqrt{3}\)
Khi x=6-3*căn 3 thì \(B=\dfrac{6-3\sqrt{3}+\sqrt{6-3\sqrt{3}}+1}{\sqrt{6-3\sqrt{3}}}\)
\(=\dfrac{7-3\sqrt{3}+\sqrt{6-3\sqrt{3}}}{\sqrt{6-3\sqrt{3}}}\)
\(=\dfrac{7\sqrt{2}-3\sqrt{6}+3-\sqrt{3}}{3-\sqrt{3}}\)