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b: \(=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8}{\sqrt{5}-1}\)
\(=2\sqrt{5}-2-2\sqrt{5}\)
=-2
c: \(=\dfrac{\sqrt{4}\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-2\sqrt{2}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)
\(=\dfrac{-3}{\sqrt{6}}=-\dfrac{\sqrt{6}}{2}\)
\(a.\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=2\sqrt{5}+\dfrac{8}{1-\sqrt{5}}=\dfrac{2\sqrt{5}-2}{1-\sqrt{5}}=\dfrac{-2\left(1-\sqrt{5}\right)}{1-\sqrt{5}}=-2\) \(b.\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=\dfrac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6.5}+\sqrt{27.6}}=\dfrac{\sqrt{2}\left(4-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{17}\right)}=-\dfrac{\sqrt{2}}{\sqrt{3}}-\dfrac{1}{\sqrt{6}}=\dfrac{-2-1}{\sqrt{6}}=-\dfrac{\sqrt{3}}{\sqrt{2}}\)
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: \(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)
Lời giải:
\(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(=\frac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{6}.\sqrt{5}+\sqrt{6}.\sqrt{27}}\)
\(=\frac{2(2\sqrt{2}-\sqrt{3})}{\sqrt{6}(\sqrt{3}-2\sqrt{2})}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{6}(\sqrt{5}+\sqrt{27})}\)
\(=\frac{-2}{\sqrt{6}}-\frac{1}{\sqrt{6}}=\frac{-3}{\sqrt{6}}\)
a) \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{2-\sqrt{3}}\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(6+2\sqrt{12}+2\right)}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(6+4\sqrt{3}+2\right)}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(8+4\sqrt{3}\right)}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\cdot4\left(2+\sqrt{3}\right)}\)
\(=\sqrt{\left(4-3\right)\cdot4}\)
\(=\sqrt{1\cdot4}\)
\(=\sqrt{4}\)
\(=2\)
b) \(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\)
\(=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)\)
\(=2\sqrt{2}+6+3\sqrt{2}+1-\left(5\sqrt{2}-7\right)\)
\(=2\sqrt{2}+6+3\sqrt{2}+1-5\sqrt{2}+7\)
\(=0+14\)
\(=14\)
c) \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
dài quá ==' cả d, e, f nữa ==' có j rảnh lm cho nhé :D
mjk đang cần gấp lắm
a: \(=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-2\left(\sqrt{5}+1\right)\)
\(=2\sqrt{5}-2\sqrt{5}-2=-2\)
c: \(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
d: \(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}+2}{2\sqrt{5}+2}=1\)
\( \dfrac{{2\sqrt 8 - \sqrt {12} }}{{\sqrt {18} - \sqrt {48} }} - \dfrac{{\sqrt 5 + \sqrt {27} }}{{\sqrt {30} - \sqrt {162} }}\\ = \dfrac{{4\sqrt 2 - 2\sqrt 3 }}{{3\sqrt 2 - 4\sqrt 3 }} - \dfrac{{\sqrt 5 + 3\sqrt 3 }}{{\sqrt {30} - 9\sqrt 2 }}\\ = \dfrac{{\left( {4\sqrt 2 - 2\sqrt 3 } \right)\left( {3\sqrt 2 + 4\sqrt 3 } \right)}}{{30}} - \dfrac{{\left( {\sqrt 5 + 3\sqrt 3 } \right)\left( {\sqrt {30} + 9\sqrt 2 } \right)}}{{ - 132}}\\ = - \dfrac{{\sqrt 6 }}{3} + \dfrac{{16\sqrt 6 + 9\sqrt {10} }}{{66}}\\ = \dfrac{{ - 22\sqrt 6 + 16\sqrt 6 + 9\sqrt {10} }}{{66}}\\ = \dfrac{{ - 6\sqrt 6 + 9\sqrt {10} }}{{66}}\\ = \dfrac{{3\left( { - 2\sqrt 6 + 3\sqrt {10} } \right)}}{{66}}\\ = \dfrac{{ - 2\sqrt 6 + 3\sqrt {10} }}{{22}} \)
\(P=\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(=\dfrac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\dfrac{\left(\sqrt{5}+\sqrt{27}\right)\left(\sqrt{30}-\sqrt{162}\right)}{\left(\sqrt{30}+\sqrt{162}\right)\left(\sqrt{30}-\sqrt{162}\right)}\)
\(=\dfrac{\sqrt{6}\left(\dfrac{4\sqrt{3}}{3}-\sqrt{2}\right)}{3\left(\sqrt{2}-\dfrac{4\sqrt{3}}{3}\right)}-\dfrac{5\sqrt{6}-9\sqrt{10}+9\sqrt{10}-27\sqrt{6}}{30-162}\)
\(=\dfrac{-\sqrt{6}}{3}-\dfrac{-22\sqrt{6}}{-132}\)
\(=\dfrac{-\sqrt{6}}{3}-\dfrac{22\sqrt{6}}{132}\)
\(=\dfrac{-44\sqrt{6}}{132}-\dfrac{22\sqrt{6}}{132}=\dfrac{-66\sqrt{6}}{132}=\dfrac{-\sqrt{6}}{2}\)
3: \(\sqrt{12-3\sqrt{7}}-\sqrt{12-3\sqrt{7}}=0\)
4: \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\)
\(=-2\sqrt{2}\)
6: \(3\sqrt{3}+4\sqrt{12}-5\sqrt{27}\)
\(=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}\)
\(=-4\sqrt{3}\)
\(a,Sửa:\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\\ =\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}\\ =2\sqrt{5}-2-2\sqrt{5}=-2\\ b,=\dfrac{\sqrt{32}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\\ =\dfrac{\sqrt{2}\left(4-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{1}{\sqrt{6}}=\dfrac{\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=\dfrac{2\sqrt{6}-\sqrt{6}}{6}=\dfrac{\sqrt{6}}{6}\)