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Ta có: \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(=\dfrac{4\sqrt{2}-3\sqrt{2}}{3\sqrt{2}-4\sqrt{2}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)
\(=-1-\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{-6-\sqrt{6}}{6}\)
\(=\dfrac{2\sqrt{8}-2\sqrt{3}}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)
\(=\dfrac{-2}{\sqrt{6}}-\dfrac{1}{\sqrt{6}}=\dfrac{-3}{\sqrt{6}}=\dfrac{-3\cdot\sqrt{6}}{6}=\dfrac{-\sqrt{6}}{2}\)
\(a,Sửa:\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\\ =\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}\\ =2\sqrt{5}-2-2\sqrt{5}=-2\\ b,=\dfrac{\sqrt{32}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\\ =\dfrac{\sqrt{2}\left(4-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{1}{\sqrt{6}}=\dfrac{\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=\dfrac{2\sqrt{6}-\sqrt{6}}{6}=\dfrac{\sqrt{6}}{6}\)
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
\(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}=\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}-3\right)^2}=\left|3-\sqrt{6}\right|+\left|\sqrt{24}-3\right|=3-\sqrt{6}+\sqrt{24}-3=2\sqrt{6}-\sqrt{6}=\sqrt{6}\)
\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=-\dfrac{\sqrt{2}\left(\sqrt{6}-4\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}=\dfrac{-\sqrt{2}}{\sqrt{3}}-\dfrac{1}{\sqrt{6}}=\dfrac{-\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{2}\).
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: \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1-\sqrt{2}-\sqrt{3}\right)\)
\(=1-\left(\sqrt{2}+\sqrt{3}\right)^2\)
\(=1-5-2\sqrt{6}\)
\(=-4-2\sqrt{6}\)
\( \dfrac{{2\sqrt 8 - \sqrt {12} }}{{\sqrt {18} - \sqrt {48} }} - \dfrac{{\sqrt 5 + \sqrt {27} }}{{\sqrt {30} - \sqrt {162} }}\\ = \dfrac{{4\sqrt 2 - 2\sqrt 3 }}{{3\sqrt 2 - 4\sqrt 3 }} - \dfrac{{\sqrt 5 + 3\sqrt 3 }}{{\sqrt {30} - 9\sqrt 2 }}\\ = \dfrac{{\left( {4\sqrt 2 - 2\sqrt 3 } \right)\left( {3\sqrt 2 + 4\sqrt 3 } \right)}}{{30}} - \dfrac{{\left( {\sqrt 5 + 3\sqrt 3 } \right)\left( {\sqrt {30} + 9\sqrt 2 } \right)}}{{ - 132}}\\ = - \dfrac{{\sqrt 6 }}{3} + \dfrac{{16\sqrt 6 + 9\sqrt {10} }}{{66}}\\ = \dfrac{{ - 22\sqrt 6 + 16\sqrt 6 + 9\sqrt {10} }}{{66}}\\ = \dfrac{{ - 6\sqrt 6 + 9\sqrt {10} }}{{66}}\\ = \dfrac{{3\left( { - 2\sqrt 6 + 3\sqrt {10} } \right)}}{{66}}\\ = \dfrac{{ - 2\sqrt 6 + 3\sqrt {10} }}{{22}} \)
\(P=\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(=\dfrac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\dfrac{\left(\sqrt{5}+\sqrt{27}\right)\left(\sqrt{30}-\sqrt{162}\right)}{\left(\sqrt{30}+\sqrt{162}\right)\left(\sqrt{30}-\sqrt{162}\right)}\)
\(=\dfrac{\sqrt{6}\left(\dfrac{4\sqrt{3}}{3}-\sqrt{2}\right)}{3\left(\sqrt{2}-\dfrac{4\sqrt{3}}{3}\right)}-\dfrac{5\sqrt{6}-9\sqrt{10}+9\sqrt{10}-27\sqrt{6}}{30-162}\)
\(=\dfrac{-\sqrt{6}}{3}-\dfrac{-22\sqrt{6}}{-132}\)
\(=\dfrac{-\sqrt{6}}{3}-\dfrac{22\sqrt{6}}{132}\)
\(=\dfrac{-44\sqrt{6}}{132}-\dfrac{22\sqrt{6}}{132}=\dfrac{-66\sqrt{6}}{132}=\dfrac{-\sqrt{6}}{2}\)
Lời giải:
\(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(=\frac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{6}.\sqrt{5}+\sqrt{6}.\sqrt{27}}\)
\(=\frac{2(2\sqrt{2}-\sqrt{3})}{\sqrt{6}(\sqrt{3}-2\sqrt{2})}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{6}(\sqrt{5}+\sqrt{27})}\)
\(=\frac{-2}{\sqrt{6}}-\frac{1}{\sqrt{6}}=\frac{-3}{\sqrt{6}}\)