\(a,=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}=\dfrac{6}{-1}=-6\\ b,=\dfrac{6\sqrt{2}+8-6\sqrt{2}+8}{\left(3\sqrt{2}-4\right)\left(3\sqrt{2}+4\right)}=\dfrac{16}{2}=8\\ c,=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\\ =\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}=\dfrac{16}{2}=8\)

\(d,=\dfrac{6\sqrt{2}+9\sqrt{3}-6\sqrt{2}+9\sqrt{3}}{\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)}=\dfrac{18\sqrt{3}}{-19}=\dfrac{-18\sqrt{3}}{19}\\ e,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\\ =\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\\ =\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)

1:

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

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

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

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

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

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

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

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

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

\(=\dfrac{2\sqrt{2}}{\sqrt{6}}\)

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

b) \(\dfrac{4\sqrt{3}-8}{2\sqrt{3}-4}-\dfrac{1}{\sqrt{5}-2}\)

\(=\dfrac{4\left(\sqrt{3}-2\right)}{2\left(\sqrt{3}-2\right)}-\dfrac{\sqrt{5}+2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)

\(=\dfrac{4}{2}-\dfrac{\sqrt{5}+2}{5-4}\)

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

\(=-\sqrt{5}\)

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

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

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

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

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

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

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

b: \(=3\sqrt{3}-\dfrac{6}{\sqrt{3}}+1-\sqrt{3}\)

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

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

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

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

\(a,\dfrac{3}{\sqrt{7}-4}+\dfrac{4+\sqrt{7}}{3}\)

\(=\dfrac{9}{3\left(\sqrt{7}-4\right)}+\dfrac{\left(\sqrt{7}-4\right)\left(\sqrt{7}+4\right)}{3\left(\sqrt{7}-4\right)}\)

\(=\dfrac{9+7-16}{3\left(\sqrt{7}-4\right)}\)

\(=0\)

\(b,\left(\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}\right):\dfrac{1}{2\sqrt{3}}\)

\(=\left[\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}\right]\cdot2\sqrt{3}\)

\(=\left(\sqrt{2}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}\right)\cdot2\sqrt{3}\)

\(=\left(\sqrt{2}+\sqrt{3}-\sqrt{2}\right)\cdot2\sqrt{3}\)

\(=\sqrt{3}\cdot2\sqrt{3}\)

\(=6\)

#\(Toru\)

\(a,\dfrac{2}{\sqrt{3}-\sqrt{5}}+\dfrac{3-2\sqrt{3}}{\sqrt{3}-2}\)

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

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

\(=\sqrt{5}+\sqrt{3}+\sqrt{3}\)

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

\(b,\dfrac{5-\sqrt{5}}{\sqrt{5}-1}+\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+3}\)

\(=\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)

\(=\sqrt{5}+\dfrac{5-2\sqrt{15}+3}{5-3}\)

\(=\dfrac{2\sqrt{5}+8-2\sqrt{15}}{2}\)

\(=\dfrac{2\cdot\left(\sqrt{5}+4-\sqrt{15}\right)}{2}\)

\(=\sqrt{5}-\sqrt{15}+4\)

#\(Toru\)

a) Ta có: \(A=\sqrt{20}-2\sqrt{45}+3\sqrt{18}+\sqrt{72}\)

\(=2\sqrt{5}-6\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)

\(=-4\sqrt{5}+15\sqrt{2}\)

b) Ta có: \(B=4\sqrt{\left(\sqrt{3}-1\right)^2}+2\sqrt{12}+4\sqrt{\dfrac{1}{2}}\)

\(=4\left(\sqrt{3}-1\right)+2\cdot2\sqrt{3}+\dfrac{4}{\sqrt{2}}\)

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

\(=8\sqrt{3}+2\sqrt{2}-4\)

c) Ta có: \(C=\left(3+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\left(3-\dfrac{3+\sqrt{3}}{1+\sqrt{3}}\right)\)

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

=9-3

=6

d) Ta có: \(D=\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\)

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

=4

a, Sửa đề:

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

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

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

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

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

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