A =  \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) +  \(\dfrac{1}{15}\) +.. .  + \(\dfrac{1}{120}\)

A =  \(\dfrac{2}{2}\).(\(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + ... + \(\dfrac{1}{120}\))

A = 2.( \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) + ... + \(\dfrac{1}{240}\))

A = 2.( \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + ... + \(\dfrac{1}{15.16}\))

A  =2 .( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{15}\) - \(\dfrac{1}{16}\))

A = 2.( \(\dfrac{1}{2}\) - \(\dfrac{1}{16}\))

A = 2.\(\dfrac{7}{16}\)

A = \(\dfrac{7}{8}\)

Ta co: A = 1/10+ 1/15+1/21+...+1/120 

              = 2/20+2/30+2/42+...+2/240=2/(4*5)+2/(5*6)+.....+2/(15*16)

              = 2*[1/(4*5)+1/(5*6)+...........+ 1/(15*16)]

              = 2* [ 1/4-1/5+1/5-1/6+.........+1/15-1/16]

              = 2*[1/4-1/16]

             = 2*3/16

              = 3/8

Vay A=3/8

 Nhân cả TS và MS các phân số của tổng với 2 thì tổng không thay đổi và ta được:

                      2/20 + 2/30 + 2/42 + 2/56 + .... + 2/240

                   = 2/4x5 + 2/5x6 + 2/6x7 + 2/7x8 + ... + 2/15x16

                   = 2 x (1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + ... + 1/15 - 1/16)

                   = 2 x (1/4 - 1/16)

                   = 2 x 3/16 = 3/8 

đúng rùi

tại sao bước thứ 4 ở trong ngoặc bạn lại lấy số đầu trừ số cuối?

1/1.4+1/4.7+1/7.10+...+1/16.19

=[1/1.4+1/4.7+1/7.10+...+1/16.19] x 3

= 3/1.4+3/4.7+3/7.10+...+3/16.19

= 1-1/4+1/4-1/7+1/7-1/10+....+1/16-1/19

=1-1/19

=18/19 :3

=6/19

ĐÂY BẠN NHÉ CHÚC BẠN HỌC TỐT NHỚ K CHO MÌNH

A = \(\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{120}\)

A= \(\dfrac{2}{2}.\left(\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{120}\right)\) 

A= \(2.\left(\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{240}\right)\)

A= \(2.\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{15.16}\right)\) 

A=\(2.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

A=\(2.\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\) 

A=\(\dfrac{2.3}{16}\) 

A= \(\dfrac{3}{8}\)

C= 1/10 + 1/15 + 1/21...........+1/120

C=1/10-1/120

C=11/12

Có: \(N=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{120}\)

\(=>N=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)

\(=>N=\frac{2}{4\cdot5}+\frac{2}{5\cdot6}+\frac{2}{6\cdot7}+...+\frac{2}{15\cdot16}\)

\(=>N=\left(\frac{2}{4}-\frac{2}{5}+\frac{2}{5}-\frac{2}{6}+...+\frac{2}{15}-\frac{2}{16}\right)\)

\(=>N=\frac{2}{4}-\frac{2}{16}\)

\(=>N=\frac{1}{2}-\frac{1}{8}\)

\(=>N=\frac{8-2}{16}=\frac{6}{16}=\frac{3}{8}\)

                            Vậy \(N=\frac{3}{8}\)

Ta có : 

\(N=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)

\(N=2\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)

\(N=2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)

\(N=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)

\(N=2\left(\frac{1}{4}-\frac{1}{16}\right)\)

\(N=\frac{1}{2}-\frac{1}{8}\)

\(N=\frac{3}{8}\)

Vậy \(N=\frac{3}{8}\)

Chúc bạn học tốt ~ 

Ta có:

\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)

\(=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)

\(=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+...+\frac{2}{15.16}\)

\(=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)

\(=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)

\(=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)

\(=2.\left(\frac{4}{16}-\frac{1}{16}\right)\)

\(=2.\frac{3}{16}=\frac{3}{8}\)

= 3/8 nhe

3/8 nhé

B = \(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{120}\)

\(B=\frac{1}{4.5:2}+\frac{1}{5.6:2}+\frac{1}{6.7:2}+.....+\frac{1}{15.16:2}\)

\(\frac{1}{2}B=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{15.16}\)

Ta thấy: \(\frac{1}{4.5}=\frac{1}{4}-\frac{1}{5};\frac{1}{5.6}=\frac{1}{5}-\frac{1}{6};\frac{1}{6.7}=\frac{1}{6}-\frac{1}{7};.....\)

\(\frac{1}{2}B=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-.....-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}\)

\(\frac{1}{2}B=\frac{1}{4}-\frac{1}{16}=\frac{3}{16}\)

\(B=\frac{3}{16}:\frac{1}{2}=\frac{3}{16}.2=\frac{3}{8}\)