\(\frac{2}{3+\sqrt{5}}+\frac{1}{2+\sqrt{5}}+\sqrt{6+2\sqrt{5}}\)
P/S:hai cái phân số để trong dấu ( )
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Những câu hỏi liên quan
AH
Akai Haruma
Giáo viên
8 tháng 7 2019
a)
\(\frac{2}{\sqrt{6}-2}+\frac{2}{\sqrt{6}+2}+\frac{5}{\sqrt{6}}=\frac{2(\sqrt{6}+2+\sqrt{6}-2)}{(\sqrt{6}-2)(\sqrt{6}+2)}+\frac{5\sqrt{6}}{6}\)
\(=\frac{4\sqrt{6}}{6-2^2}+\frac{5\sqrt{6}}{6}=2\sqrt{6}+\frac{5\sqrt{6}}{6}=\frac{17\sqrt{6}}{6}\)
b)
\(\frac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}-\frac{1}{\sqrt{3}+\sqrt{2}+\sqrt{5}}=\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}-(\sqrt{3}+\sqrt{2}-\sqrt{5})}{(\sqrt{3}+\sqrt{2}-\sqrt{5})(\sqrt{3}+\sqrt{2}+\sqrt{5})}\)
\(=\frac{2\sqrt{5}}{(\sqrt{3}+\sqrt{2})^2-5}=\frac{2\sqrt{5}}{5+2\sqrt{6}-5}=\sqrt{\frac{5}{6}}\)
AH
Akai Haruma
Giáo viên
8 tháng 7 2019
c)
\(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right):\frac{1}{\sqrt{5}-\sqrt{2}}\)
\(=\left[\frac{\sqrt{2}(\sqrt{3}-1)}{1-\sqrt{3}}-\sqrt{5}\right].(\sqrt{5}-\sqrt{2})\)
\(=(-\sqrt{2}-\sqrt{5})(\sqrt{5}-\sqrt{2})=-(\sqrt{5}+\sqrt{2})(\sqrt{5}-\sqrt{2})\)
\(=-(5-2)=-3\)
d)
\(\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)
\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{1}{4}+\frac{2}{2\sqrt{6}}+\frac{1}{6}}\)
\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{(\frac{1}{2}-\frac{1}{\sqrt{6}})^2}\)
\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}(\frac{1}{2}-\frac{1}{\sqrt{6}})\)
\(=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{2\sqrt{3}}-\frac{1}{3\sqrt{2}}=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}\)
S
3 tháng 7 2019
\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\left(-\sqrt{7}-\sqrt{5}\right):\frac{1}{\sqrt{7}-\sqrt{5}}=\frac{\sqrt{5}-\sqrt{7}}{\sqrt{7}+\sqrt{5}}=\frac{\left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{7}+\sqrt{5}\right)^2}=\frac{2}{12+2\sqrt{35}}\)
S
3 tháng 7 2019
\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+3\right)}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{8-2\sqrt{15}}{2}+\frac{8+2\sqrt{15}}{2}-\frac{\left(\sqrt{5}+1\right)^2}{4}=8-\frac{6+2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}\)
27 tháng 8 2020
a) Ta có: \(\frac{\sqrt{5}-2}{5+2\sqrt{5}}-\frac{1}{2+\sqrt{5}}+\frac{1}{\sqrt{5}}\)
\(=\frac{\sqrt{5}-2}{\sqrt{5}\left(\sqrt{5}+2\right)}-\frac{\sqrt{5}}{\sqrt{5}\left(\sqrt{5}+2\right)}+\frac{\sqrt{5}+2}{\sqrt{5}\left(\sqrt{5}+2\right)}\)
\(=\frac{\sqrt{5}-2-\sqrt{5}+\sqrt{5}+2}{\sqrt{5}\left(\sqrt{5}+2\right)}\)
\(=\frac{\sqrt{5}}{\sqrt{5}\left(\sqrt{5}+2\right)}\)
\(=\frac{1}{\sqrt{5}+2}\)
b) Ta có: \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
\(=\frac{\sqrt{6}\left(\sqrt{3}+1\right)}{\sqrt{6}\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}+\frac{\sqrt{2}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}{\sqrt{6}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}-\frac{2\sqrt{2}\cdot\left(\sqrt{3}+2\right)}{\sqrt{6}\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}\)
\(=\frac{3\sqrt{2}+\sqrt{6}+\sqrt{2}\cdot\left(5+3\sqrt{3}\right)-2\sqrt{6}-4\sqrt{2}}{\sqrt{6}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}\)
\(=\frac{-\sqrt{2}-\sqrt{6}+5\sqrt{2}+3\sqrt{6}}{\sqrt{6}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}\)
\(=\frac{4\sqrt{2}+2\sqrt{6}}{\sqrt{6}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}\)
\(=\frac{2\sqrt{2}\cdot\left(2+\sqrt{3}\right)}{\sqrt{3}\cdot\sqrt{2}\cdot\left(2+\sqrt{3}\right)\left(\sqrt{3}+1\right)}\)
\(=\frac{2}{3+\sqrt{3}}\)