Đặt tổng trân là A

Ta có A=1/2x3+1/3x4+...+1/10x11

A=1/2-1/3+1/3-1/4+...+1/10-1/11

Sau khi rút gọn ta được A=1/2-1/11

A=9/22

Nguyễn Trung Hiếu nó copy

Đặt A = 1/2 + 5/6 + ... + 89/90

9 - A = 1- 1/2 + 1- 5/6 + ... + 1- 89/90

 9 -A = 1/2 + 1/6 + ... + 1/90

9 -A = 1/1.2 + 1/2.3 + ... + 1/9.10

9-A = 1/1 - 1/2 + 1/2 - 1/3 +....  + 1/9 - 1/10

9 - A = 1/1 - 1/10

9- A = 9/10

=> A = 9 - 9/10

=> A = 81/10 

số cần tìm là 81/10

A=1/2+1/6+1/12+1/20+...+1/72+1/90

A=1/1x2+1/2x3+1/3x4+1/4x5+.....+1/8x9+1/9x10

A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+....+1/8-1/9+1/9-1/10

A=1-1/10

A=9/10

k cho mik nha

A=1/(1*2)+1/(2*3)+1/(3*4)+.....+1/(9*10)

=1/1-1/2+1/2-1/3+.......+1/9-1/10

=1/1-1/10=9/10

\(\frac{81}{10}\)

bằng -9/10

\(=\frac{-9}{10}\)

C = -1/90 - 1/72 - 1/56 - 1/42 - 1/30 - 1/20 - 1/12 -1/6 - 1/2  

C = -1/90 - ( 1/72 + 1/56 + 1/42 + 1/30 + 1/20 + 1/12+1/6 +1/2 ) ( áp dụng quy tắc dấu ngoặc ở lớp 6 ) 

C = -1/90 - [ 1/ ( 9 . 8 ) + 1/ ( 8 . 7 ) + 1/ ( 6 . 7 ) + ...+ 1/(3 .4 ) + 1/ ( 2 . 3 ) + 1/ ( 1. 2 ) ] 

C =  -1/90 - (     1 - 1/2 + 1/2 -1 /3 + .. + 1/7 - 1/8 + 1/8 - 1/9 ) 

C =  -1/90 - ( 1 - 1/9 ) 

C =  -1/90 - 8/9 

C = -1/90 - 80/90

C =   -81/90

C =   -9/10 

Chúc bạn học giỏi 

\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{90}+\dfrac{1}{110}\)

\(=\dfrac{1}{2x3}+\dfrac{1}{3x4}+\dfrac{1}{4x5}+...+\dfrac{1}{9x10}+\dfrac{1}{10x11}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\)

\(=\dfrac{1}{2}-\dfrac{1}{11}=\dfrac{11}{22}-\dfrac{2}{22}=\dfrac{9}{22}\)

1/6+1/12+1/20+1/90+1/110

=1/2x3+1/3x4+1/4x5+...+1/9x10+1/10x11

=1/2-1/3+1/3-1/4+1/4-1/5+1/5-...+1/9-1/10+1/10-1/11

=1/2-1/11=9/22

\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)

\(=\dfrac{9}{10}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

\(=\dfrac{9}{10}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)

\(=\dfrac{9}{10}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=\dfrac{9}{10}-\left(1-\dfrac{1}{10}\right)=\dfrac{9}{10}-\dfrac{9}{10}=0\)

\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-...-\dfrac{1}{6}-\dfrac{1}{2}=-\left(-\dfrac{9}{10}+\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+...+\dfrac{1}{6}+\dfrac{1}{2}\right)\)

\(=-\left(-\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)

\(=-\left(-\dfrac{9}{10}+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)

\(=-\left(-\dfrac{9}{10}+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=-\left(-\dfrac{9}{10}+1-\dfrac{1}{10}\right)=-\left(-\dfrac{9}{10}+\dfrac{9}{10}\right)=0\)

Ta có :

\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)

\(=\dfrac{9}{10}-\left(\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}+\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{2}\right)\)

\(=\dfrac{9}{10}-\left(\dfrac{1}{9.10}+\dfrac{1}{9.8}+\dfrac{1}{7.6}+\dfrac{1}{6.5}+\dfrac{1}{5.4}+\dfrac{1}{4.3}+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)

\(=\dfrac{9}{10}-\left(\dfrac{1}{10}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{8}+..........+\dfrac{1}{2}-\dfrac{1}{1}\right)\)

\(=\dfrac{9}{10}-\left(1-\dfrac{1}{10}\right)\)

\(=\dfrac{9}{10}-\dfrac{9}{10}=0\)

\(\text{1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90}\)

\(\text{= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9 + 1/9.1}\)

\(\text{= 1/1 - 1/2 + 1/2 - 1/3 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/6 - 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 - 1/10}\)

\(=1/1-1/10-10/10-1/10-9/10\)

Vậy \(\text{ 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 + 1/90 = 9/10}\)

Sửa đề:

\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)

\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\dfrac{1}{2}-\dfrac{1}{10}\)

\(=\dfrac{3}{5}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

\(=1-\frac{1}{11}\)

\(=\frac{10}{11}\)

10/11 nha

chúc bạn học tốt tk nha ^.^

chúc bạn học tốt tk nha ^.^

   \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+......+\frac{1}{90}+\frac{1}{110}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

 \(=1-\left(\frac{1}{2}-\frac{1}{2}-\frac{1}{3}-\frac{1}{3}-....-\frac{1}{10}-\frac{1}{10}\right)\frac{1}{11}\)

 \(=1-\frac{1}{11}\)

 \(=\frac{10}{11}\)