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

\(A=\dfrac{2\sqrt{3}}{\sqrt{3}+1}+3\sqrt{\dfrac{1}{6}}\cdot\sqrt{\dfrac{1}{2}}-\sqrt{12}\)

\(A=\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}+3\cdot\dfrac{1}{\sqrt{6}}\cdot\dfrac{1}{\sqrt{2}}-2\sqrt{3}\)

\(A=\dfrac{2\sqrt{3}\cdot\left(\sqrt{3}-1\right)}{2}+3\cdot\dfrac{1}{\sqrt{12}}-2\sqrt{3}\)

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

\(A=3-\sqrt{3}+\dfrac{3}{2\sqrt{3}}-2\sqrt{3}\)

\(A=3-3\sqrt{3}+\dfrac{\sqrt{3}}{2}\)

\(A=\dfrac{6+6\sqrt{3}+\sqrt{3}}{2}\)

\(A=\dfrac{6+7\sqrt{3}}{2}\)

Lời giải:
\(\frac{1}{1-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}=\frac{1+\sqrt{2}}{(1-\sqrt{2})(1+\sqrt{2})}-\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{4}}{(\sqrt{3}-\sqrt{4})(\sqrt{3}+\sqrt{4})}\)

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

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

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

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

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

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

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

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

=-3

giúp mình với ạ 

b: Ta có: \(\dfrac{4}{\sqrt{3}+1}+\dfrac{2}{\sqrt{3}-1}-\dfrac{6}{3-\sqrt{3}}\)

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

\(=2\sqrt{3}-4\)

Ta có: \(\sqrt{12}-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}+\dfrac{11}{2\sqrt{3}+1}\)

\(=2\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}+\dfrac{11\left(2\sqrt{3}-1\right)}{\left(2\sqrt{3}+1\right)\left(2\sqrt{3}-1\right)}\)

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

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

\(=3\sqrt{3}-1\)

Ta có : \(\sqrt{12}-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}+\dfrac{11}{2\sqrt{3}+1}\)

\(=\sqrt{12}-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}+\dfrac{\left(2\sqrt{3}+1\right)\left(2\sqrt{3}-1\right)}{2\sqrt{3}+1}\)

\(=\sqrt{12}-\sqrt{3}+2\sqrt{3}-1=2\sqrt{3}-\sqrt{3}+2\sqrt{3}-1\)

\(=3\sqrt{3}-1\)

 b) \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)

\(=\dfrac{\sqrt{2}\cdot\sqrt{12-3\sqrt{7}}-\sqrt{2}\cdot\sqrt{12+3\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(\sqrt{21}\right)^2-2\cdot\sqrt{21}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{21}\right)^2+2\cdot\sqrt{21}\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{21}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{21}+\sqrt{3}\right)^2}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}\)

\(=\dfrac{-2\sqrt{3}}{\sqrt{2}}\)

\(=-\sqrt{6}\)  

c) \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}\)

\(=\sqrt[3]{\dfrac{3\cdot9}{4\cdot16}}\)

\(=\sqrt[3]{\left(\dfrac{3}{4}\right)^3}\)

\(=\dfrac{3}{4}\)

d) \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}\)

\(=\sqrt[3]{\dfrac{54}{-2}}\)

\(=\sqrt[3]{-27}\)

\(=\sqrt[3]{\left(-3\right)^3}\)

\(=-3\) 

a: Sửa đề: \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)

\(=\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}\cdot\sqrt{6}}+\dfrac{\sqrt{3}-\sqrt{2}}{12}\)

\(=\dfrac{\sqrt{6}+1}{3\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{12}\)

\(=\dfrac{2\sqrt{2}\left(\sqrt{6}+1\right)+\sqrt{3}-\sqrt{2}}{12}\)

\(=\dfrac{4\sqrt{3}+2\sqrt{2}+\sqrt{3}-\sqrt{2}}{12}\)

\(=\dfrac{5\sqrt{3}+\sqrt{2}}{12}\)

e: \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

\(=\sqrt[3]{2\sqrt{2}+3\sqrt{2}+6+1}-\sqrt[3]{2\sqrt{2}-3\sqrt{2}+6-1}\)

\(=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}\)

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

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

\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=\sqrt{100}-\sqrt{1}=10-1=9\)

cả 2 ý bạn trục căn thức ở mấu là xong nhé:

vd: \(\dfrac{1}{\sqrt{1}+\sqrt{2}}=\dfrac{\sqrt{1}-\sqrt{2}}{-1}\). Rồi tương tự như vậy