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9 tháng 10 2019

 \(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+....+\frac{1}{2020}\left(1+2+3+...+2020\right)\)

\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{2020}.\frac{2020.2021}{2}\)

\(=1+\frac{3}{2}+\frac{4}{2}+....+\frac{2021}{2}\)

\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+....+\frac{2021}{2}\)

\(=\frac{\left[\left(2021-2\right)+1\right]\left(2021+2\right)}{2}:2\)

\(=1021615\)

20 tháng 7 2018

\(A=\left(-2\right)\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right).\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{214}\right)\)

\(=2.\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{215}{214}=215\)

\(B=\left(-1\frac{1}{2}\right).\left(-1\frac{1}{3}\right).\left(-1\frac{1}{4}\right)....\left(-1\frac{1}{299}\right)\)

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{300}{299}=\frac{300}{2}=150\)

\(C=-\frac{7}{4}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{333333}{424242}\right)\)

\(=-\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

\(=-\frac{7}{4}.33.\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

\(=-\frac{231}{4}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(=-\frac{231}{4}\left(\frac{1}{3}-\frac{1}{7}\right)\)

\(=-\frac{231}{4}.\frac{4}{21}=-11\)

8 tháng 2 2017

31/29

8 tháng 2 2017

\(=\frac{-\frac{1}{8}-\frac{27}{64}.4}{-2+\frac{9}{16}-\frac{3}{8}}\)

\(=\frac{-\frac{1}{8}-\frac{27}{16.4}.4}{-2+\frac{9-6}{16}}\)

\(=\frac{-\frac{1}{8}-\frac{27}{16}}{-2+\frac{3}{16}}\)

\(=\frac{-\left(\frac{2+27}{16}\right)}{\frac{-32+3}{16}}\)

\(=\frac{-\frac{29}{16}}{\frac{-29}{16}}\)

\(=1\)

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a)

\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A =  - 1\end{array}\)

b)

\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A =  - 1 + 0 + 0 =  - 1\end{array}\)

26 tháng 3 2017

\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2017^2}\right)\)

\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{2016.2018}{2017^2}\)

\(=\frac{2.3^2.4^2.5^2...2016^2.2017.2018}{2^2.3^2.4^2.5^2...2017^2}\)

\(=\frac{2018}{2.2017}=\frac{1009}{2017}\)

3 tháng 2 2017

lơp 6  ko bt