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TT
0
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 6 2019
a/ \(\frac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\frac{8\left(1+\sqrt{5}\right)}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)
b/ \(\frac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}=\frac{-2}{\sqrt{6}}-\frac{1}{\sqrt{6}}=\frac{-3}{\sqrt{6}}=-\frac{\sqrt{6}}{2}\)
c/ \(\frac{\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}+\frac{\sqrt{\left(2+\sqrt{3}\right)^2}}{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}=2-\sqrt{3}+2+\sqrt{3}=4\)
d/ \(\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\frac{\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\frac{\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)}{8}=\frac{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{4}=1\)
e/ \(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\frac{\sqrt{2}}{2+\sqrt{4+2\sqrt{3}}}+\frac{\sqrt{2}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{\sqrt{2}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{\sqrt{2}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}=\frac{\sqrt{2}}{3+\sqrt{3}}+\frac{\sqrt{2}}{3-\sqrt{3}}=\frac{\sqrt{2}\left(3-\sqrt{3}+3+\sqrt{3}\right)}{6}=\sqrt{2}\)
f/ \(\frac{9+4\sqrt{5}-8\sqrt{5}}{2\left(\sqrt{5}-2\right)}=\frac{9-4\sqrt{5}}{2\left(\sqrt{5}-2\right)}=\frac{\left(\sqrt{5}-2\right)^2}{2\left(\sqrt{5}-2\right)}=\frac{\sqrt{5}-2}{2}\)
28 tháng 8 2020
a) Ta có: \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\)
\(=\frac{\left(10+2\sqrt{10}\right)\left(1-\sqrt{5}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(1-\sqrt{5}\right)}+\frac{8\left(\sqrt{5}+\sqrt{2}\right)}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
\(=\frac{10-10\sqrt{5}+2\sqrt{10}-10\sqrt{2}+8\sqrt{5}+8\sqrt{2}}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
\(=\frac{10-2\sqrt{5}+2\sqrt{10}-2\sqrt{2}}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
\(=\frac{2\sqrt{5}\left(\sqrt{5}-1\right)+2\sqrt{2}\left(\sqrt{5}-1\right)}{-\left(\sqrt{5}-1\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
\(=\frac{2\cdot\left(\sqrt{5}-1\right)\left(\sqrt{5}+\sqrt{2}\right)}{-\left(\sqrt{5}-1\right)\left(\sqrt{5}+\sqrt{2}\right)}\)
\(=\frac{2}{-1}=-2\)
b) Ta có: \(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
\(=\frac{-2\left(\sqrt{3}-\sqrt{8}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)
\(=\frac{-2}{\sqrt{6}}-\frac{1}{\sqrt{6}}\)
\(=-\frac{3}{\sqrt{6}}=\frac{-\sqrt{3}}{\sqrt{2}}\)
c) Ta có: \(\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}\)
\(=\sqrt{\frac{\left(2-\sqrt{3}\right)\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}+\sqrt{\frac{\left(2+\sqrt{3}\right)\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)
\(=\sqrt{\frac{7-4\sqrt{3}}{4-3}}+\sqrt{\frac{7+4\sqrt{3}}{4-3}}\)
\(=\sqrt{4-2\cdot2\cdot\sqrt{3}+3}+\sqrt{4+2\cdot2\cdot\sqrt{3}+3}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left|2-\sqrt{3}\right|+\left|2+\sqrt{3}\right|\)
\(=2-\sqrt{3}+2+\sqrt{3}\)(Vì \(2>\sqrt{3}>0\))
\(=4\)
d) Ta có: \(\frac{\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
\(=\frac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\frac{\sqrt{5-2\cdot\sqrt{5}\cdot1+1}\cdot\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\left(6+2\sqrt{5}\right)}{4\left(\sqrt{5}+1\right)}\)
\(=\frac{\left|\sqrt{5}-1\right|\cdot\left(5+2\cdot\sqrt{5}\cdot1+1\right)}{2\cdot\left(\sqrt{5}+1\right)\cdot2}\)
\(=\frac{\left(\sqrt{5}-1\right)\cdot\left(\sqrt{5}+1\right)^2}{2\cdot\left(\sqrt{5}+1\right)\cdot2}\)(Vì \(\sqrt{5}>1\))
\(=\frac{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}{4}\)
\(=\frac{5-1}{4}=\frac{4}{4}=1\)
e) Ta có: \(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\frac{\sqrt{2}-\sqrt{2+\sqrt{3}}}{\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2}-\sqrt{2+\sqrt{3}}\right)}+\frac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2}+\sqrt{2-\sqrt{3}}\right)}\)
\(=\frac{\sqrt{2}-\sqrt{2+\sqrt{3}}}{2-\left(2+\sqrt{3}\right)}+\frac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{2-\left(2-\sqrt{3}\right)}\)
\(=\frac{2-\sqrt{4+2\sqrt{3}}}{\sqrt{2}\cdot\left(2-2-\sqrt{3}\right)}+\frac{2+\sqrt{4-2\sqrt{3}}}{\sqrt{2}\cdot\left(2-2+\sqrt{3}\right)}\)
\(=\frac{2-\sqrt{3+2\cdot\sqrt{3}\cdot1+1}}{-\sqrt{6}}+\frac{2+\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{6}}\)
\(=\frac{-2+\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{6}}+\frac{2+\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{6}}\)
\(=\frac{\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|}{\sqrt{6}}\)
\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{6}}\)
\(=\frac{2\sqrt{3}}{\sqrt{6}}=\frac{\sqrt{12}}{\sqrt{6}}=\sqrt{2}\)
f) Ta có: \(\frac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}\)
\(=\frac{9+4\sqrt{5}-8\sqrt{5}}{2\left(\sqrt{5}-2\right)}\)
\(=\frac{9-4\sqrt{5}}{2\cdot\left(\sqrt{5}-2\right)}\)
\(=\frac{5-2\cdot\sqrt{5}\cdot2+2}{2\cdot\left(\sqrt{5}-2\right)}\)
\(=\frac{\left(\sqrt{5}-2\right)^2}{2\left(\sqrt{5}-2\right)}\)
\(=\frac{\sqrt{5}-2}{2}\)
NH
1
12 tháng 8 2021
Ta có: \(A=\left(\sqrt{6}+\sqrt{10}\right)\cdot\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
=5-3=2