\(\frac{1}{21}\)+ \(\frac{1}{22}\)
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10 tháng 5 2020

\(N=\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+\frac{1}{24}+\frac{1}{25}+\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+\frac{1}{29}+\frac{1}{30}\) >\(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+\frac{1}{30}=\frac{1}{30}.10=\frac{1}{3}\)

=> N > \(\frac{1}{3}\)

23 tháng 4 2017

ta có 1/3=10/30

1/21+1/22+...+1/30 có 10 p/số

mà 1/21>1/30

1/22>1/30

....

1/29>1/30

1/30=1/30

=>1/21+..1/30>1/30+....1/30 có 10 phân số 

=>1/21+...1/30>1/3

23 tháng 4 2017

Ta có: \(\frac{1}{21}< \frac{1}{30}\)

\(\frac{1}{22}< \frac{1}{30}\)

......

\(\frac{1}{29}< \frac{1}{30}\)

\(\Rightarrow S< \frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\)(có 10 p/s)
\(\Rightarrow S< \frac{1}{30}.10=\frac{10}{30}=\frac{1}{3}\)

Vậy S < 1/3

3 tháng 8 2020

a) \(\frac{7}{5}.\frac{-31}{125}.\frac{1}{2}.\frac{10}{17}.\frac{-1}{2^3}=\frac{7.\left(-31\right).1.10.\left(-1\right)}{5.2.125.17.2^3}=\frac{31.7}{17.125.2^3}=\frac{217}{17000}\)

b) \(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).0=0\)

c) \(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{3.4.5...100}{2.3.4...99}=\frac{100}{2}=50\)

d) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-99}{100}=\frac{-\left(1.2.3..99\right)}{2.3.4...100}=-\frac{1}{100}\)

e) \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}=\frac{\left(1.2.3..29\right).\left(3.4.5...31\right)}{\left(2.3.4...30\right).\left(2.3.4...30\right)}\)

\(=\frac{1.31}{30.2}=\frac{31}{60}\)

23 tháng 6 2017

\(A=\frac{8}{9}\cdot\frac{15}{16}\cdot\frac{24}{25}\cdot...\cdot\frac{360}{361}\cdot\frac{399}{400}\)

\(A=\frac{2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot...\cdot18\cdot20\cdot19\cdot21}{3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot19\cdot19\cdot20\cdot20}\)

\(A=\frac{2\cdot21}{3\cdot20}\)

\(A=\frac{7}{10}\)

\(B=\frac{9}{8}\cdot\frac{16}{15}\cdot\frac{25}{24}\cdot...\cdot\frac{441}{440}\cdot\frac{484}{483}\)

\(B=\frac{3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot21\cdot21\cdot22\cdot22}{2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot...\cdot20\cdot22\cdot21\cdot23}\)

\(B=\frac{3\cdot22}{2\cdot23}=\frac{33}{23}\)

\(C=\frac{17}{23}.\left(\frac{7}{61}+\frac{28}{61}+\frac{26}{61}\right)\)

\(C=\frac{17}{23}\cdot1=\frac{17}{23}\)

5 tháng 5 2019

\(\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot.....\cdot\frac{899}{30^2}\)

\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot.....\cdot\frac{29\cdot31}{30\cdot30}\)

\(=\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{3}{4}\cdot\frac{5}{4}\cdot....\cdot\frac{29}{30}\cdot\frac{31}{30}\)

\(=\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{29}{30}\right)\cdot\left(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot....\cdot\frac{31}{30}\right)\)

\(=\frac{1}{30}\cdot\frac{31}{2}\)

\(=\frac{31}{60}\)

b, \(A=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)

Ta có:

\(\frac{3}{15}< \frac{3}{10}=\frac{3}{10}\)

\(\frac{3}{15}< \frac{3}{11}< \frac{3}{10}\)

\(\frac{3}{15}< \frac{3}{12}< \frac{3}{10}\)

\(\frac{3}{15}< \frac{3}{13}< \frac{3}{10}\)

\(\frac{3}{15}< \frac{3}{14}< \frac{3}{10}\)

\(\Rightarrow\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}< \frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}\)

\(\Rightarrow\frac{3\cdot5}{15}< A< \frac{3\cdot5}{10}\)

\(\Rightarrow1< A< \frac{15}{10}=\frac{3}{2}\)

\(\frac{3}{2}< 2\)

\(\Rightarrow1< A< 2\)

c ,Ta có

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{25}\right)+\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)

5 tháng 5 2019

thanks!!!vui

30 tháng 4 2018

a)\(=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}\)

\(=\left(\frac{-3}{7}+\frac{3}{7}\right)-\left(\frac{15}{26}+\frac{2}{13}\right)\)

\(=0-\frac{19}{26}\)

\(=-\frac{19}{26}\)

30 tháng 4 2018

c)\(=\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)

\(=\frac{-11}{23}.2-\frac{1}{23}\)

\(=\frac{-22}{23}-\frac{1}{23}\)

\(=-1\)