Số cần tìm là:

 A = 1/10 = 0,1

 k nha

A = \(\frac{1}{10}=0,1\)

Tích cho tớ để đủ 250 điểm đi 

=> A = 1/6*7 + 1/7*8 + 1/8*9 + 1/9*10 + ... + 1/13*14 + 1/14*15                                                                                    = 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10 +...+ 1/13 - 1/14 + 1/14 - 1/15                                                        = 1/6 - 1/15                                                                                                                                                            =1/10                Nhớ ^^

=1/6*7+1/7*8+1/8*9...+1/14*15

=1/6-1/7+1/7-1/8+...+1/14-1/15

=1/6-1/15

=1/10

1/42 + 1/56 + 1/72 + 1/90 + 1/110 + 1/132 + 1/156 + 1/182 + 1/210

= 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10 + 1/10.11 + 1/11.12 + 1/12.13 + 1/13.14 + 1/14.15

= 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10 + 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13 + 1/13 - 1/14 + 1/14 - 1/15

= 1/6 - 1/15

= 1/10

\(=\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{14.15}\)

\(=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{15}\)

\(=\frac{1}{6}-\frac{1}{15}\)

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

\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+\frac{1}{182}+\frac{1}{210}\)

\(A=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}\)

\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\)

\(A=\frac{1}{6}-\frac{1}{15}\)

\(A=\frac{1}{10}\)

A=\(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+\frac{1}{182}+\frac{1}{210}\)

=\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}\)

=\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{14}-\frac{1}{15}\)

=\(\frac{1}{6}-\frac{1}{15}=\frac{1}{10}\)

\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+...+\frac{1}{210}=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+...+\frac{1}{14.15}\)

\(=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+...+\frac{1}{14}-\frac{1}{15}\)

\(=\frac{1}{6}-\frac{1}{15}=\frac{1}{10}\)

\(A=\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+\frac{1}{182}+\frac{1}{210}\)

\(A=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}\)

\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\)

\(A=\frac{1}{6}-\frac{1}{15}\)

\(A=\frac{1}{10}\)

\(\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+....+\frac{1}{240}=\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+...+\frac{1}{15.16}\)

\(=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+...+\frac{1}{15}-\frac{1}{16}=\frac{1}{7}-\frac{1}{16}=\frac{9}{112}\)

\(=\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+\frac{1}{13.14}+\frac{1}{14.15}+\frac{1}{15.16}\)

\(=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{16}\)

\(=\frac{1}{7}-\frac{1}{16}\)

\(=\frac{9}{112}\)

Tách:

56 = 7 x 8

72 = 8 x 9

90 = 9 x 10

110 = 10 x 11

132 = 11 x 12

156 = 12 x 13

182 = 13 x 14

210 = 14 x 15

240 = 15 x 16

=> \(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\frac{1}{12\cdot13}+\frac{1}{13\cdot14}+\frac{1}{14\cdot15}+\frac{1}{15\cdot16}\)

Đựợc dãy sai phân, làm tiếp được

\(\frac{1}{7}-\frac{1}{16}=\frac{9}{112}\)

Vậy kết quả là \(\frac{9}{112}\)

Bài 1: Ta có:

\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\)

\(=\dfrac{20}{8.14}+\dfrac{20}{14.20}+\dfrac{20}{20.26}+\dfrac{20}{26.32}\)

\(=\dfrac{20}{6}\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+\dfrac{6}{26.32}\right)\)

\(=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+...+\dfrac{1}{26}-\dfrac{1}{32}\right)\)

\(=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{32}\right)=\dfrac{20}{6}.\dfrac{3}{32}=\dfrac{5}{16}\)

Vậy \(M=\dfrac{5}{16}\)

Bài 2: Ta có:

\(A=\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+...+\dfrac{1}{210}\)

\(=\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+...+\dfrac{1}{14.15}\)

\(=\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{14}-\dfrac{1}{15}\)

\(=\dfrac{1}{6}-\dfrac{1}{15}=\dfrac{1}{10}\)

Vậy \(A=\dfrac{1}{10}\)

Giải:

\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}.\)

\(M=\dfrac{5}{28}+\dfrac{5}{70}+\dfrac{5}{130}+\dfrac{5}{208}.\)

\(M=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+\dfrac{5}{13.16}.\)

\(M=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right).\)

\(M=\dfrac{5}{3}\left[\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+\left(\dfrac{1}{10}-\dfrac{1}{10}\right)+\left(\dfrac{1}{13}-\dfrac{1}{13}\right)+\left(\dfrac{1}{4}-\dfrac{1}{16}\right)\right].\)

\(M=\dfrac{5}{3}\left[0+0+0+\left(\dfrac{1}{4}-\dfrac{1}{16}\right).\right]\)

\(M=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{16}\right).\)

\(M=\dfrac{5}{3}\left(\dfrac{4}{16}-\dfrac{1}{16}\right).\)

\(M=\dfrac{5}{3}.\dfrac{3}{16}.\)

\(M=\dfrac{15}{48}=\dfrac{5}{16}.\)

M=\(\dfrac{5}{28}+\dfrac{1}{14}+\dfrac{1}{26}+\dfrac{5}{208}\)

M=\((\dfrac{5}{28}+\dfrac{1}{14})+\left(\dfrac{1}{26}+\dfrac{5}{208}\right)\)

M=\(\dfrac{1}{4}+\dfrac{1}{16}\)

M=\(\dfrac{5}{16}\)

A=\(\dfrac{1}{6\times7}+\dfrac{1}{7\times8}+\dfrac{1}{8\times9}+...+\dfrac{1}{13\times14}+\dfrac{1}{14\times15}\)

A=\(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{14}-\dfrac{1}{15}\)

Sau khi giản ước các phân số cho nhau Ta có:

A=\(\dfrac{1}{6}-\dfrac{1}{15}\)

A=\(\dfrac{1}{10}\)

Ta có:\(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{210}\) 

   \(=\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+....+\frac{1}{14.15}\)

    \(=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{15}\)

      \(=\frac{1}{6}-\frac{1}{15}=\frac{1}{10}\)

\(\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}+\dfrac{1}{156}+\dfrac{1}{182}+\dfrac{1}{210}\\ =\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}+\dfrac{1}{13.14}+\dfrac{1}{14.15}\\ =\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{13}-\dfrac{1}{14}\\ =\dfrac{1}{5}-\dfrac{1}{14}\\ =\dfrac{14}{70}-\dfrac{5}{70}=\dfrac{9}{70}\)

\(=\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{12.13}\)

áp dụng \(\frac{1}{a.b}=\frac{1}{a}-\frac{1}{b}\)làm sẽ có các số nghịch đảo và được kết quả là 1/4 - 1/13

A = 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 + 1/110 + 1/132 + 1/156

A = 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.10 + 1/10.11 + 1/11.12 + 1/12.13

A = 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10 + 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13

A = 1/4 - 1/13

A = 9/52

A = \(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{132}+\frac{1}{156}\)

    = \(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{11.12}+\frac{1}{12.13}\)

    = \(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}\)

    = \(\frac{1}{4}-\frac{1}{13}\)

    = \(\frac{9}{52}\)

Vậy \(A=\frac{9}{52}\)