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26 tháng 9 2016

\(\left(\frac{\frac{17}{24}.9\frac{1}{2}-3\frac{1}{4}.\frac{17}{24}}{3\frac{1}{2}.2\frac{13}{36}+2\frac{13}{36}.2\frac{3}{4}}-\frac{1}{2}\right)^{-2}\)

\(=\left(\frac{\frac{17}{24}.\left(9\frac{1}{2}-3\frac{1}{4}\right)}{2\frac{13}{36}.\left(3\frac{1}{2}+2\frac{3}{4}\right)}-\frac{1}{2}\right)^{-2}\)

\(=\left(\frac{\frac{17}{24}.\left(\frac{19}{2}-\frac{13}{4}\right)}{\frac{85}{36}.\left(\frac{7}{2}+\frac{11}{4}\right)}-\frac{1}{2}\right)^{-2}\)

\(=\left(\frac{\frac{17}{24}.\frac{19.2-13}{4}}{\frac{85}{36}.\frac{7.2+11}{4}}-\frac{1}{2}\right)^{-2}\)

\(=\left(\frac{\frac{17}{24}.\frac{25}{4}}{\frac{85}{36}.\frac{25}{4}}-\frac{1}{2}\right)^{-2}\)

\(=\left(\frac{17}{24}:\frac{85}{36}-\frac{1}{2}\right)^{-2}\)

\(=\left(\frac{17}{24}.\frac{36}{85}-\frac{1}{2}\right)^{-2}\)

\(=\left(\frac{3}{10}-\frac{1}{2}\right)^{-2}\)

\(=\left(\frac{3-5}{10}\right)^{-2}\)

\(=\left(\frac{-1}{5}\right)^{-2}\)

\(=\frac{1}{\left(-\frac{1}{5}\right)^2}=\frac{1}{\frac{\left(-1\right)^2}{5^2}}=\frac{1}{\frac{1}{25}}=25\)

9 tháng 8 2017

\(\frac{\frac{1}{4}+\frac{1}{24}+\frac{1}{124}}{\frac{3}{4}+\frac{3}{24}+\frac{3}{124}}+\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{127}}{\frac{3}{7}+\frac{3}{17}+\frac{3}{127}}=\frac{\frac{1}{4}+\frac{1}{24}+\frac{1}{124}}{3\left(\frac{1}{4}+\frac{1}{24}+\frac{1}{124}\right)}+\frac{2\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{127}\right)}{3\left(\frac{1}{7}+\frac{1}{17}+127\right)}=\frac{1}{3}+\frac{2}{3}=\) \(1\)

\(1\frac{1}{30}:\left(24\frac{1}{6}-24\frac{1}{5}\right)-\frac{1\frac{1}{2}-\frac{3}{4}}{4x-\frac{1}{2}}=\left(-1\frac{1}{15}\right):\left(8\frac{1}{5}-8\frac{1}{3}\right)\)

\(\Rightarrow\frac{31}{30}:\left(\frac{145}{6}-\frac{121}{5}\right)-\frac{\frac{3}{2}-\frac{3}{4}}{4x-\frac{1}{2}}=\left(\frac{-16}{15}\right):\left(\frac{41}{5}-\frac{25}{3}\right)\)

\(\Rightarrow\frac{31}{30}:\left(\frac{725}{30}-\frac{726}{30}\right)-\frac{\frac{6}{4}-\frac{3}{4}}{4x-\frac{1}{2}}=\left(\frac{-16}{15}\right):\left(\frac{123}{15}-\frac{125}{15}\right)\)

\(\Rightarrow\frac{31}{30}:\frac{-1}{30}-\frac{\frac{3}{4}}{4x-\frac{1}{2}}=\frac{-16}{15}:\frac{-2}{15}\)

\(\Rightarrow\frac{31}{30}.\frac{30}{-1}-\frac{3}{4}:\left(4x-\frac{1}{2}\right)=\frac{-16}{15}.\frac{15}{-2}\)

\(\Rightarrow\left(-31\right)-\frac{3}{4}:\left(4x-\frac{1}{2}\right)=8\)

\(\Rightarrow\frac{3}{4}:\left(4x-\frac{1}{2}\right)=\left(-31\right)-8\)

\(\Rightarrow\frac{3}{4}:\left(4x-\frac{1}{2}\right)=\left(-39\right)\)

\(\Rightarrow4x-\frac{1}{2}=\frac{3}{4}:\left(-39\right)\)

\(\Rightarrow4x-\frac{1}{2}=\frac{3}{4}.\frac{-1}{39}\)

\(\Rightarrow4x-\frac{1}{2}=\frac{-1}{52}\)

\(\Rightarrow4x=\frac{-1}{52}+\frac{1}{2}\)

\(\Rightarrow4x=\frac{-1}{52}+\frac{26}{52}\)

\(\Rightarrow4x=\frac{25}{52}\)

\(\Rightarrow x=\frac{25}{52}:4\)

\(\Rightarrow x=\frac{25}{52}.\frac{1}{4}\)

\(\Rightarrow x=\frac{25}{208}\)

Vậy \(x=\frac{25}{208}\)

Chúc bn học tốt

6 tháng 3 2020

thank you

NV
13 tháng 3 2020

\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

\(\Rightarrow\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+25\sqrt{24}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=1-\frac{1}{5}=\frac{4}{5}\)

7 tháng 1 2017

\(U\left(n\right)=\frac{1}{\left(n+1\right).\sqrt{n}+n\sqrt{n+1}}\)

\(U\left(n\right)=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n.\left(n+1\right)^2-n^2\left(n+1\right)}=\frac{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)}{n\left(n+1\right)\left(n+1-n\right)}\)

\(U\left(n\right)=\frac{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n}\sqrt{n+1}\right)^2}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

\(S_{u\left(n\right)}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=1-\frac{1}{5}< 1\)