\(\frac{12}{3-\sqrt{3}}\)
K
Khách
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8 tháng 9 2017
Ta có : \(\frac{3}{\sqrt{n}+\sqrt{n+4}}=\frac{3}{4}.\frac{4}{\sqrt{n}+\sqrt{n+4}}=\frac{3}{4}.\frac{4\left(\sqrt{n+4}-\sqrt{n}\right)}{\left(\sqrt{n+4}+\sqrt{n}\right)\left(\sqrt{n+4}-\sqrt{n}\right)}\)
\(=\frac{3}{4}.\frac{4\left(\sqrt{n+4}-\sqrt{n}\right)}{n+4-n}=\frac{3}{4}.\frac{4\left(\sqrt{n+4}-\sqrt{n}\right)}{4}=\frac{3}{4}\left(\sqrt{n+4}-\sqrt{n}\right)\)
Áp dụng ta được :
\(\frac{3}{\sqrt{4}+\sqrt{8}}+\frac{3}{\sqrt{8}+\sqrt{12}}+\frac{3}{\sqrt{12}+\sqrt{16}}+...+\frac{3}{\sqrt{572}+\sqrt{576}}\)
\(=\frac{3}{4}\left(\sqrt{8}-\sqrt{4}+\sqrt{12}-\sqrt{8}+\sqrt{16}-\sqrt{12}+...+\sqrt{576}-\sqrt{572}\right)\)
\(=\frac{3}{4}\left(\sqrt{576}-\sqrt{4}\right)=\frac{3}{4}\left(24-4\right)=\frac{3}{4}.20=15\)
L
0
25 tháng 6 2016
B=\(\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}=\frac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\frac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
C=\(\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}=\frac{3\left(1+\sqrt{3}\right)}{\sqrt{3}}+\frac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}=\sqrt{3}+1-\sqrt{3}=1\)
D=\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
E=\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\sqrt{3}+\frac{1}{2-\sqrt{3}}=\frac{2\sqrt{3}-1}{2-\sqrt{3}}\)
BA
1
9 tháng 8 2020
\(\frac{2}{\sqrt{3}}+\frac{\sqrt{2}}{3}+\frac{2}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}=\frac{2\sqrt{3}}{3}+\frac{\sqrt{2}}{3}+\frac{2\sqrt{3}}{3}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)
\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\sqrt{12}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\sqrt{12\left(\frac{5}{12}-\frac{1}{\sqrt{6}}\right)}\)
\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\sqrt{5-2\sqrt{6}}=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\cdot\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\left|\sqrt{3}-\sqrt{2}\right|=\frac{2\sqrt{3}+\sqrt{2}}{3}+\frac{1}{3}\left(\sqrt{3}-\sqrt{2}\right)\)(vì \(\sqrt{3}-\sqrt{2}>0\))
\(=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{3}=\sqrt{3}\)
Bạn không ghi rõ đề nên tớ sẽ làm như này
\(\frac{12}{3-\sqrt{3}}\)
\(=\frac{36+12\sqrt{5}}{3^2-\sqrt{5^2}}\)
\(=\frac{12\left(3+\sqrt{5}\right)}{3^2-\sqrt{5^2}}\)
\(=\frac{12\left(3+\sqrt{5}\right)}{9-5}\)
\(=\frac{12\left(3+\sqrt{5}\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)\)
Sosuke bạn làm sai rồi kìa. dò lại đi