\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)

\(\dfrac{1}{\sqrt{3}+\sqrt{2}+1}=\dfrac{\sqrt{3}-\sqrt{2}-1}{\left(\sqrt{3}+\sqrt{2}+1\right)\left(\sqrt{3}-\sqrt{2}-1\right)}\)

\(=\dfrac{\sqrt{3}-\sqrt{2}-1}{3-\left(\sqrt{2}+1\right)^2}=\dfrac{\sqrt{3}-\sqrt{2}-1}{-2\sqrt{2}}=\dfrac{\left(\sqrt{3}-\sqrt{2}-1\right)\sqrt{2}}{-2\sqrt{2}.\sqrt{2}}=\dfrac{\sqrt{6}-2-\sqrt{2}}{-4}\)

\(=\dfrac{2+\sqrt{2}-\sqrt{6}}{4}\)

\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)

\(\dfrac{1}{\sqrt{3}+\sqrt{2}+1}=\dfrac{2+\sqrt{2}-\sqrt{6}}{4}\)

1) Ta có: \(3\sqrt{12}+\dfrac{1}{2}\sqrt{48}-\sqrt{27}\)

\(=3\cdot2\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-3\sqrt{3}\)

\(=6\sqrt{3}+2\sqrt{3}-3\sqrt{3}\)

\(=5\sqrt{3}\)

2) Ta có: \(\dfrac{2}{\sqrt{3}-5}\)

\(=\dfrac{2\left(\sqrt{3}+5\right)}{\left(\sqrt{3}-5\right)\left(\sqrt{3}+5\right)}\)

\(=\dfrac{2\left(\sqrt{3}+5\right)}{3-25}\)

\(=\dfrac{-2\left(\sqrt{3}+5\right)}{22}\)

\(=\dfrac{-\sqrt{3}-5}{11}\)

3) Ta có: \(\sqrt{\dfrac{2}{5}}\)

\(=\dfrac{\sqrt{2}}{\sqrt{5}}\)

\(=\dfrac{\sqrt{2}\cdot\sqrt{5}}{5}\)

\(=\dfrac{\sqrt{10}}{5}\)

\(\dfrac{1}{\sqrt{a}-1}\)

\(=\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

\(=\dfrac{\sqrt{a}+1}{\left(\sqrt{a}\right)^2-1^2}\)

\(=\dfrac{\sqrt{a}+1}{a-1}\)

\(\frac{1}{1+\sqrt{2}+\sqrt{3}}\)

\(=\frac{1+\sqrt{2}-\sqrt{3}}{\left(1+\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)}\)

\(=\frac{1+\sqrt{2}-\sqrt{3}}{2\sqrt{2}}\)

\(=\frac{2+\sqrt{2}-\sqrt{6}}{4}\)

bạn hãy nhân ở mẫu với biểu thức tương ướng để tạo ra biểu thức liên hợp , là HĐT số 3 ạ 

\(\dfrac{1}{1-\sqrt{2}}=-1-\sqrt{2}\)