\(\sqrt{\frac{1}{16}}\)= ?
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25 tháng 11 2015
= 0,6 : 5/4 + 1/4 + 2/9 : 5/9 - 1/4
= 3/5 . 4/5 + 2/9 . 9/5
= 12/25 + 2/5
= 22/25
TN
4
13 tháng 8 2016
\(\left(\frac{2}{5}\sqrt{16}+2\sqrt{\frac{16}{25}}\right):2\sqrt{\frac{1}{16}}=\left(\frac{2}{5}.\sqrt{4^2}+2\sqrt{\frac{4^2}{5^2}}\right):\frac{2}{\sqrt{4^2}}\)
\(=\left(\frac{2}{5}.4+2.\frac{4}{5}\right).2=\left(\frac{8}{5}+\frac{8}{5}\right).2=\frac{32}{5}\)
13 tháng 8 2016
\(\left(\frac{2}{5}.\sqrt{16}+2\sqrt{\frac{16}{25}}\right):2\sqrt{\frac{1}{16}}\)
\(=\left(\frac{2}{5}.4+2.\frac{4}{5}\right):2.\frac{1}{4}\)
\(=\left(\frac{8}{5}+\frac{8}{5}\right):\frac{1}{2}\)
\(=\frac{16}{5}:\frac{1}{2}\)
\(=\frac{32}{5}\)
^...^ 
^_^
8 tháng 8 2017
Với n > 0 Ta có:
\(\frac{1}{\sqrt{n+1}-\sqrt{n}}=\frac{\sqrt{n+1}+\sqrt{n}}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}\)
\(=\sqrt{n+1}+\sqrt{n}\)
\(\Rightarrow\frac{1}{\sqrt{16}-\sqrt{15}}-\frac{1}{\sqrt{15}-\sqrt{14}}+...+\frac{1}{\sqrt{10}-\sqrt{9}}\)
\(=\sqrt{16}+\sqrt{15}-\sqrt{15}-\sqrt{14}+...+\sqrt{10}+\sqrt{9}\)
\(\sqrt{16}+\sqrt{9}=3+4=7\)
NC
5 tháng 10 2020
Ta có:
\(A=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\cdot\sqrt{\frac{5}{12}-\frac{1}{16}}\)
\(A=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\cdot\sqrt{\frac{17}{48}}\)
\(A=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{51}}{12}\)
\(A=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{17}}{12}\)
\(A=\frac{4\sqrt{3}+2\sqrt{2}+\sqrt{17}}{12}\)
5 tháng 10 2020
Ta có: \(\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}=\sqrt{\frac{5}{12}-\frac{\sqrt{6}}{6}}=\sqrt{\frac{5-2\sqrt{6}}{12}}\)
Vì \(5-2\sqrt{6}=3-2\sqrt{3}.\sqrt{2}+2=\left(\sqrt{3}\right)^2-2\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2\)\(\Rightarrow5-2\sqrt{6}=\left(\sqrt{3}-\sqrt{2}\right)^2\)
Như vậy: \(\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}=\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}=\frac{1}{2\sqrt{3}}\left(\sqrt{3}-\sqrt{2}\right)\)
Lại có: \(\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{1}{\sqrt{3}}.\frac{1}{2\sqrt{3}}\left(\sqrt{3}-\sqrt{2}\right)\)
Rút gọn ta được \(A=\frac{\sqrt{3}}{2}\)
TH
3
23 tháng 7 2018
\(\left(\sqrt{1\frac{9}{16}}-\sqrt{\frac{9}{16}}\right)\)
\(=\left(\sqrt{\frac{25}{16}}-\sqrt{\frac{9}{16}}\right)\)
\(=\left(\sqrt{\left(\frac{5}{4}\right)^2}-\sqrt{\left(\frac{3}{4}\right)^2}\right)\)
\(=\left(\frac{5}{4}-\frac{3}{4}\right)=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}\)
...Vậy ...................
23 tháng 7 2018
\(\sqrt{1\frac{9}{16}}-\sqrt{\frac{9}{16}}=\sqrt{\frac{25}{16}}-\sqrt{\frac{9}{16}}\)
\(=\sqrt{\left(\frac{5}{4}\right)^2}-\sqrt{\left(\frac{3}{4}\right)^2}=\frac{5}{4}-\frac{3}{4}=\frac{2}{4}=\frac{1}{2}\)
#
25 tháng 8 2019
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{15}+\sqrt{16}}\)
= \(-\left(\sqrt{1}-\sqrt{2}\right)-\left(\sqrt{2}-\sqrt{3}\right)-...-\left(\sqrt{15}-\sqrt{16}\right)\)
=\(-\left(\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+...+\sqrt{15}-\sqrt{16}\right)\)
=\(-\left(1-\sqrt{16}\right)=-\left(1-4\right)=3\)
1/4 nha
\(\sqrt{\frac{1}{6}}=\frac{\sqrt{6}}{6}=0,4082482905\)