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tổng trên sẽ là:

1/1+1/3+1/210=3/321

đáp số:3/321

9 tháng 8 2017

Đặt : \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\)

\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)

\(2A=1-\frac{1}{101}\)

\(2A=\frac{100}{101}\)

\(A=\frac{100}{101}\cdot\frac{1}{2}=\frac{50}{101}\)

9 tháng 8 2017

Ta có:

a)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)

b)

 \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{210}\)

\(=2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}\right)\)

\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\right)\)

\(=2.\left(\frac{1}{2}-\frac{1}{21}\right)=2.\frac{19}{42}=\frac{19}{21}\)

18 tháng 7 2023

A = \(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\)+...+\(\dfrac{1}{182}\)+ 210

 A = \(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\)+\(\dfrac{1}{3\times4}\)\(\dfrac{1}{4\times5}\)+...+ \(\dfrac{1}{13\times14}\)+ 210

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)\(\dfrac{1}{3}\)\(\dfrac{1}{4}\)\(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+...+\(\dfrac{1}{13}\) - \(\dfrac{1}{14}\) + 210

A = 1 - \(\dfrac{1}{14}\) + 210

A = 211 - \(\dfrac{1}{14}\)

A = \(\dfrac{2953}{14}\)

4 tháng 10 2016

: A = 1/6+1/12+1/20+1/30+.........+1/210

A = 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + ... + 1/14.15

A = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/14 - 1/15

A = 1/2 - 1/15

A = 13/30

4 tháng 10 2016

Ta có: 1/2= 1/1- 1/2 
1/6= 1/2 - 1/3 
1/12= 1/3- 1/4 
... 
1/30= 1/5 - 1/6 
1/42= 1/6 - 1/7 
Thay vào tổng kia: 1/2+1/6+...+1/30+1/42= 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/5 - 1/6 + 1/6 - 1/7 = 1/2 - 1/7= 5/14 
Chúc bạn học tốt. Thân!

25 tháng 7 2018

\(B=\frac{3}{1}+\frac{3}{3}+\frac{3}{6}+...+\frac{3}{210}\)

\(=\frac{6}{2}+\frac{6}{6}+\frac{6}{12}+...+\frac{6}{420}\)

\(=\frac{6}{1.2}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{20.21}\)

\(=6\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\right)\)

\(=6\left(1-\frac{1}{21}\right)\)

\(=6.\frac{20}{21}=\frac{40}{7}\)

\(25\frac{2}{13}-\left(\frac{15}{17}+15\frac{2}{3}\right)=\frac{327}{13}-\left(\frac{15}{17}+\frac{47}{3}\right)\)

\(=\frac{327}{13}-\frac{844}{51}\)

\(=\frac{5705}{663}\)

\(\frac{5}{30}+\frac{15}{90}+\frac{250}{150}+\frac{350}{210}+\frac{45}{270}=\frac{1}{6}+\frac{1}{6}+\frac{5}{3}+\frac{5}{3}+\frac{1}{6}\)

\(=\frac{3}{6}+\frac{10}{3}\)

\(=\frac{1}{2}+\frac{10}{3}\)

\(=\frac{23}{6}\)

\(3\frac{1}{11}.\frac{27}{46}.1\frac{6}{17}.2\frac{4}{9}=\frac{34}{11}.\frac{27}{46}.\frac{23}{17}.\frac{22}{9}\)

\(=\frac{68}{9}.\frac{27}{34}\)

\(=6\)

13 tháng 10 2019

a. x : 1/7 + x:36/7 = 1/3

=> 7x + 7x/36 = 1/3

=> 7x(1 + 1/36) = 1/3

=> 7x.37/36 = 1/3

=> 7x = 12/37

=> x = 12/259

b, tương tự nhá

3 tháng 12 2017

      \(\frac{7}{10}-\frac{3}{10}x\frac{1}{10}+\frac{4}{25}\)

\(=\frac{7}{10}-\frac{3}{100}+\frac{4}{25}\)

\(=\frac{70}{100}-\frac{3}{100}+\frac{4}{25}\)

\(=\frac{67}{100}+\frac{4}{25}\)

\(=\frac{67}{100}+\frac{16}{100}\)

\(=\frac{83}{100}=0,83\)

      \(\frac{6}{11}x\frac{3}{7}+\frac{3}{7}x\frac{5}{11}\)

\(=\frac{18}{77}+\frac{15}{77}\)

\(=\frac{33}{77}=\frac{3}{7}\)

     \(49x\left(311-4807:23\right)+98210\)

\(=49x\left(311-209\right)+98210\)

\(=49x102+98210\)

\(=4998+98210\)

\(=103208\)

3 tháng 12 2017

a) \(\frac{7}{10}-\frac{3}{10}\times\frac{1}{10}+\frac{4}{25}\)

\(=\frac{2}{5}\times\frac{1}{10}+\frac{4}{25}\)

\(=\frac{1}{25}+\frac{4}{25}\)

\(=\frac{5}{25}=\frac{1}{5}\)

b)\(\frac{6}{11}\times\frac{3}{7}+\frac{3}{7}\times\frac{5}{11}\)

\(=\frac{3}{7}\times\left(\frac{6}{11}+\frac{5}{11}\right)\)

\(=\frac{3}{7}\times1\)

\(=\frac{3}{7}\)

c)\(49\times\left(311-4807:23\right)+98210\)

\(=49\times\left(311-209\right)+98210\)

\(=49\times102+98210\)

\(=4998+98210\)

\(=103208\)