Rút gọn biểu thức \(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\), ta có kết quả là
\(\dfrac{3-\sqrt{3}}{3}\). \(\dfrac{3+\sqrt{3}}{3}\). \(2\sqrt{3}\). \(3-\sqrt{3}\). Hướng dẫn giải:\(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)
\(=\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{\sqrt{3}}-\dfrac{2}{\sqrt{3}\left(\sqrt{3}+1\right)}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)+\left(2+\sqrt{3}\right)\left(\sqrt{3}+1\right)-2\left(2+\sqrt{3}\right)}{\sqrt{3}.\left(2+\sqrt{3}\right)\left(\sqrt{3}+1\right)}\)
\(=\dfrac{2\sqrt{3}+4}{\sqrt{3}.\left(2+\sqrt{3}\right).\left(\sqrt{3}+1\right)}\)
\(=\dfrac{2\left(\sqrt{3}+2\right)}{\sqrt{3}\left(2+\sqrt{3}\right)\left(\sqrt{3}+1\right)}\)
\(=\dfrac{2}{\sqrt{3}.\left(\sqrt{3}+1\right)}\)
\(=\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{3.\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{3.2}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{3}=\dfrac{3-\sqrt{3}}{3}\)
