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15 tháng 8 2018

\(\frac{3}{10\cdot13}+\frac{3}{13\cdot16}+...+\frac{3}{58\cdot61}\)

\(=\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+...+\frac{1}{58}-\frac{1}{61}\)

\(=\frac{1}{10}-\frac{1}{61}\)

\(=\frac{51}{610}\)

15 tháng 8 2018

\(\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}+...+\frac{3}{58.61}\)

\(=\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+...+\frac{1}{58}-\frac{1}{61}\)

\(=\frac{1}{10}-\frac{1}{61}\)

\(=\frac{51}{610}\)

Học tốt !

18 tháng 5 2016

\(A=\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+...+\frac{1}{58}-\frac{1}{61}\)

\(A=\frac{1}{10}-\frac{1}{61}=\frac{51}{610}\)

18 tháng 5 2016

A=3/10.13 +3/13.16+ 3/16.19+....+3/58.61

A=1/10.13+1/13.16+1/16.19+.....+1/58.61

A=1/10- 1/13+ 1/13- 1/16+ 1/16- 1/19+...+1/58 –1/61

A=1/10 – 1/61

A= 61/610 – 10/610

A= 51/610

Mình giải xong rồi k nhá?

20 tháng 3 2020

(3/2+3/8) +( 3/4 =3/16) +( 3/32+ 3/128)+(3,64+3,256)+3/512

= 4+6/2+6/16+6/896+3/512

=10/2+13/056+3/512

=23/256+3/512

=26/768

18 tháng 7 2018

3 câu như nhau cả thôi :v

\(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{55\cdot57}\)

\(A=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{55\cdot57}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{55}-\frac{1}{57}\right)\)

\(A=\frac{1}{2}\left(1-\frac{1}{57}\right)\)

\(A=\frac{1}{2}\cdot\frac{56}{57}\)

\(A=\frac{28}{57}\)

28 tháng 9 2021

\(1,\\ \dfrac{3}{5}=\dfrac{21}{35};\dfrac{4}{7}=\dfrac{20}{35}\\ \dfrac{2}{3}=\dfrac{4}{6};\dfrac{5}{6}=\dfrac{5}{6}\\ \dfrac{4}{9}=\dfrac{8}{18};\dfrac{1}{6}=\dfrac{3}{18}\\ 2,\\ \dfrac{5}{9}=\dfrac{10}{18};\dfrac{7}{8}=\dfrac{14}{16};\dfrac{24}{42}=\dfrac{12}{21}\)

28 tháng 9 2021

1. \(\dfrac{3}{5}=\dfrac{21}{35};\dfrac{4}{7}=\dfrac{20}{35}\)

\(\dfrac{2}{3}=\dfrac{4}{6};\dfrac{5}{6}=\dfrac{5}{6}\)

\(\dfrac{4}{9}=\dfrac{16}{36};\dfrac{1}{6}=\dfrac{6}{36}\)

 

2. \(\dfrac{5}{9}=\dfrac{10}{18}\)

\(\dfrac{7}{8}=\dfrac{14}{16}\)

\(\dfrac{24}{42}=\dfrac{12}{21}\)

 

C = 3/4.7 + 3/7.10 + 3/10.13 + ... + 3/73.76

=1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + ... + 1/73 - 1/76

=1/4 - 1/76

=18/76

16 tháng 7 2016

\(C=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+......+\frac{1}{73}-\frac{1}{76}\)

\(=\frac{1}{4}-\frac{1}{76}\)

\(=\frac{19}{76}-\frac{1}{76}\)

\(=\frac{18}{76}=\frac{9}{38}\)

20 tháng 2 2020

2,997070313

\(\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\frac{3}{16}+\frac{3}{32}+\frac{3}{64}+\frac{3}{128}+\frac{3}{256}+\frac{3}{512}+\frac{3}{1024}\)

=\(3.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}+\frac{1}{1024}\right)\)

=\(3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}-\frac{1}{128}+\frac{1}{128}-\frac{1}{256}+\frac{1}{256}-\frac{1}{512}+\frac{1}{512}-\frac{1}{1024}\right)\)

=\(3.\left(1-\frac{1}{1024}\right)=3.\left(\frac{1024}{1024}-\frac{1}{1024}\right)=3.\frac{1023}{1024}=\frac{3069}{1024}\)

Chúc em học tốt

3 tháng 6 2018

\(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+.....+\frac{3}{1+2+...+100}\)

     \(=3+\frac{3}{3}+\frac{3}{6}+...+\frac{3}{5050}\)

        \(=\frac{2}{2}.\left(3+\frac{3}{3}+\frac{3}{6}+...+\frac{3}{5050}\right)\)

          \(=\frac{6}{2}+\frac{6}{6}+\frac{6}{12}+...+\frac{6}{10100}\)

          \(=6.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\right)\)

            \(=6.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)

             \(=6.\left(1-\frac{1}{101}\right)\)

               \(=6.\frac{100}{101}=\frac{600}{101}\)

Vậy \(A=\frac{600}{101}\)

3 tháng 6 2018

\(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+...+100}\)

\(A=\frac{3.2}{2}+\frac{3.2}{\left(1+2\right).2}+\frac{3.2}{\left(1+2+3\right).2}+...+\frac{3.2}{\left(1+2+...+100\right).2}\)

\(A=\frac{6}{2}+\frac{6}{6}+\frac{6}{12}+...+\frac{6}{10100}\)

\(A=\frac{6}{1.2}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{100.101}\)

\(A=6\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)

\(A=6\cdot\left(1-\frac{1}{101}\right)=6\cdot\frac{100}{101}=\frac{600}{101}\)

Vay A = ........