1/20 + 1/30 + 1/42 + ... + 1/156

= 1/4.5 + 1/5.6 + 1/6.7 + .... + 1/12.13

= 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/12 - 1/13

= 1/4 - 1/13

= 9/52

\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}\)

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

\(=\frac{1}{4}-\frac{1}{13}=\frac{9}{52}\)

**** 

Đặt tổng trên là A ta có :

 A= 1/ 20 + 1/ 30 + 1/ 42 + 1/ 56 + 1/ 72 + 1/90 + 1/110 + 1 / 123 + 1/ 156

    =  1 / 4 x5 + 1/ 5 x 6 + 1/6x 7 + 1/ 7x8 + 1/8x9 + 1/9x10+ 1/ 10x11+ 1 /11x12 +1/12 x 13

      = 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

       = 1 /4 - 1 /13

        = 9 /52
 

\(B=\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}+\dfrac{1}{156}\)

\(B=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}+\dfrac{1}{12\cdot13}\)

\(B=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}\)

\(B=\dfrac{1}{5}-\dfrac{1}{13}=\dfrac{8}{65}\)

B=\(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}\)

B=\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-.......-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)

B=\(\dfrac{1}{5}-\dfrac{1}{12}=\dfrac{7}{60}\)

(sở dĩ có như trên vì \(\dfrac{1}{x\left(x+1\right)}=\dfrac{1}{x}-\dfrac{1}{x+1}\) nên cứ áp dụng vào là ra  nhé)

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

B = 1/5x6 + 1/6x7 + 1/7x8 + 1/8x9 + 1/9x10 + 1/10x11 + 1/11x12 + 1/12x13

B = 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

B = 1/5 - 1/13 

B = 8/65

Ta viết lại biểu thức A như sau:

\(A=-\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{11.12}\right)\)

\(A=-\left(\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+...+\dfrac{12-11}{11.12}\right)\)

\(A=-\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{12}\right)\)

\(A=-\left(\dfrac{1}{4}-\dfrac{1}{12}\right)\)

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

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 

\(\dfrac{1}{20}=\dfrac{1}{4x5}=\dfrac{1}{4}-\dfrac{1}{5}\)

Tương tự các phân số khác

S= \(\dfrac{1}{4}-\dfrac{1}{12}=\dfrac{1}{6}\)

\(\dfrac{1}{20}+\dfrac{1}{30}\)+ \(\dfrac{1}{42}\)+\(\dfrac{1}{56}\)+\(\dfrac{1}{72}\)+\(\dfrac{1}{90}\)+\(\dfrac{1}{110}\)+\(\dfrac{1}{132}\)

= \(\dfrac{1}{4\times5}\)+\(\dfrac{1}{5\times6}\)+\(\dfrac{1}{6\times7}\)+\(\dfrac{1}{7\times8}\)+\(\dfrac{1}{8\times9}\)+\(\dfrac{1}{9\times10}\)+\(\dfrac{1}{10\times11}\)+\(\dfrac{1}{11\times12}\)

= \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+\(\dfrac{1}{9}\)-\(\dfrac{1}{10}\)+\(\dfrac{1}{10}\)-\(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{12}\)

= \(\dfrac{1}{4}\) - \(\dfrac{1}{12}\)

= \(\dfrac{3}{12}\) - \(\dfrac{1}{12}\)

= \(\dfrac{2}{12}\)

=\(\dfrac{1}{6}\)

`=1/[4xx5]+1/[5xx6]+1/[6xx7]+...+1/[11xx12]`

`=1/4-1/5+1/5-1/6+1/6-1/7+...+1/11-1/12`

`=1/4-1/12=3/12-1/12=2/12=1/6`

\(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\\ =\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}+\dfrac{1}{8\times9}+\dfrac{1}{9\times10}+\dfrac{1}{10\times11}+\dfrac{1}{11\times12}\\ =\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\\ =\dfrac{1}{4}-\dfrac{1}{12}\\ =\dfrac{3}{12}-\dfrac{1}{12}=\dfrac{2}{12}=\dfrac{1}{6}\)