a) Ta có: \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)

\(=\dfrac{-2\left(\sqrt{3}-\sqrt{8}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{6}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)

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

b) Ta có: \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1-\sqrt{2}-\sqrt{3}\right)\)

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

\(=1-5-2\sqrt{6}\)

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

Mysterious Person giúp e với! Em cảm ơn!!!

a: \(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=4\sqrt{5}\)

b: \(=2\sqrt{5}-2-2\sqrt{5}=-2\)

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

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

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

e: \(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)

a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)

b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)

c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)

d: \(=\sqrt{5}\left(\sqrt{3}-1\right)-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)

=căn 5-1/2*căn 5

=1/2*căn 5

e: \(=\dfrac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\dfrac{1}{\sqrt{6}}=\dfrac{2}{\sqrt{6}}-\dfrac{1}{\sqrt{6}}=\dfrac{1}{\sqrt{6}}\)

f:=2+căn 3+căn 2-2-căn 3=căn 2

\(a,Sửa:\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\\ =\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}\\ =2\sqrt{5}-2-2\sqrt{5}=-2\\ b,=\dfrac{\sqrt{32}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\\ =\dfrac{\sqrt{2}\left(4-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{1}{\sqrt{6}}=\dfrac{\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=\dfrac{2\sqrt{6}-\sqrt{6}}{6}=\dfrac{\sqrt{6}}{6}\)

Ta có: \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)

\(=\dfrac{4\sqrt{2}-3\sqrt{2}}{3\sqrt{2}-4\sqrt{2}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)

\(=-1-\dfrac{1}{\sqrt{6}}\)

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

\(=\dfrac{2\sqrt{8}-2\sqrt{3}}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)

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

\(=\frac{3\sqrt{10}-2\sqrt{6}}{22}\)

\( \dfrac{{2\sqrt 8 - \sqrt {12} }}{{\sqrt {18} - \sqrt {48} }} - \dfrac{{\sqrt 5 + \sqrt {27} }}{{\sqrt {30} - \sqrt {162} }}\\ = \dfrac{{4\sqrt 2 - 2\sqrt 3 }}{{3\sqrt 2 - 4\sqrt 3 }} - \dfrac{{\sqrt 5 + 3\sqrt 3 }}{{\sqrt {30} - 9\sqrt 2 }}\\ = \dfrac{{\left( {4\sqrt 2 - 2\sqrt 3 } \right)\left( {3\sqrt 2 + 4\sqrt 3 } \right)}}{{30}} - \dfrac{{\left( {\sqrt 5 + 3\sqrt 3 } \right)\left( {\sqrt {30} + 9\sqrt 2 } \right)}}{{ - 132}}\\ = - \dfrac{{\sqrt 6 }}{3} + \dfrac{{16\sqrt 6 + 9\sqrt {10} }}{{66}}\\ = \dfrac{{ - 22\sqrt 6 + 16\sqrt 6 + 9\sqrt {10} }}{{66}}\\ = \dfrac{{ - 6\sqrt 6 + 9\sqrt {10} }}{{66}}\\ = \dfrac{{3\left( { - 2\sqrt 6 + 3\sqrt {10} } \right)}}{{66}}\\ = \dfrac{{ - 2\sqrt 6 + 3\sqrt {10} }}{{22}} \)