Câu hỏi của Lê Thúy Hằng - Toán lớp 6 | Học trực tuyến

\(B=\frac{2.5-1}{2.5}+\frac{5.8-1}{5.8}+\frac{8.11-1}{8.11}+...+\frac{50.53-1}{50.53}\)

\(B=1-\frac{1}{2.5}+1-\frac{1}{5.8}+1-\frac{1}{8.11}+...+1-\frac{1}{50.53}\)

\(B=17-\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{50.53}\right)\)

\(B=17-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{50}-\frac{1}{53}\right)\)

\(B=17-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{53}\right)=\frac{1785}{106}\)

nhân 3 vào cả hai vế 

\(A=\frac{5}{2.5}+\frac{5}{5.8}+\frac{5}{8.11}+...+\frac{5}{98.101}\)

\(=\frac{5}{2}-\frac{5}{5}+\frac{5}{5}-\frac{5}{8}+....+\frac{5}{98}-\frac{5}{101}\)

\(=\frac{5}{2}-\frac{5}{101}=\frac{495}{202}\)

\(\frac{5}{2\times5}+\frac{5}{5\times8}+\frac{5}{8\times11}+...+\frac{5}{98\times101}\)

\(=\frac{5}{3}\times\left(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{98\times101}\right)\)

\(=\frac{5}{3}\times\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)

\(=\frac{5}{3}\times\left(\frac{1}{2}-\frac{1}{101}\right)\)

\(=\frac{5}{3}\times\frac{99}{202}=\frac{165}{202}\)

18a

19b

20a

18:A

19:B

20:A

18 a 19 b 20 a

=> 3B = 3.( 1/2.5 + 1/5.8 + 1/8.11 + ........... + 1/122.125)

           = 3/2.5 + 3/5.8 + 3/ 8.11 + ......+ 3/122.125

Ta có: 3/ 2.5 = 1/2 - 1/5 

          3/5.8  = 1/5 -1/8

          3/ 8.11 = 1/8 -1/11

          ..........................

         3/122 . 125 = 3/122 - 3/125

=> 3B=  1/2 - 15/5 + 1/5 -1/8 +1/8 - 1/11 +........+1/122 - 1/125

         =  1/2 - 1/125 = 125/250 - 2/250= 123/250

=> B= 3B : 3 = 123/250 :3 = 123/250 . 1/3 = 41/250

=> 2C = 2.(1/9.11 + 1/11.13 +....+ 1/97 .99)

           = 2/9.11 + 2/11 .13 +.....+ 2/ 97.99

Ta có: 2/9.11 = 1/9 - 1/11

          2/11.13 = 2/11 -2/ 13

         ...............................

         2/97.99 = 1/97 - 1/99

=> 2B = 1/9 - 1/11 + 1/11 - 1/13 + ....+ 1/97 - 1/99

           = 1/9 -1/99 = 11/99 - 1/99 =10/99

=> B= 2B : B = 10/99 :2 =10/99 . 1/2 = 5/99

Vậy B = 5/99

 \(B=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{122.125}\)

\(3B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{122.125}\)

Nhận xét:

\(\frac{3}{2.5}=\frac{1}{2}-\frac{1}{5}=\frac{3}{10}\)

\(\frac{3}{5.8}=\frac{1}{5}-\frac{1}{8}=\frac{3}{40}\)

\(\frac{3}{8.11}=\frac{1}{8}-\frac{1}{11}=\frac{3}{88}\)

.............

Từ nhận xét trên ta có:

\(3B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{122}-\frac{1}{125}\)

\(3B=\frac{1}{2}-\frac{1}{125}=\frac{123}{250}\)

\(B=\frac{123}{250}:3=\frac{41}{250}\)

\(C=\frac{1}{9.11}+\frac{1}{11.13}+...+\frac{1}{97.99}\)

\(2C=\frac{2}{9.11}+\frac{2}{11.13}+...+\frac{2}{97.99}\)

Nhận xét:

\(\frac{2}{9.11}=\frac{1}{9}-\frac{1}{11}=\frac{2}{99}\)

\(\frac{2}{11.13}=\frac{1}{11}-\frac{1}{13}=\frac{1}{143}\)

..................

Từ nhận xét trên ta có:

\(2C=\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{99}\)

\(2C=\frac{1}{9}-\frac{1}{99}=\frac{10}{99}\)

\(C=\frac{10}{99}:2=\frac{5}{99}\)

1/2.5 + 1/5.8 + ... + 1/49.52

= 1/3 . ( 3/2.5 + 3/5.8 + ... + 3/49.52 )

= 1/3 . ( 1/2 - 1/5 + 1/5 - 1/8 + ... + 1/49 - 1/52 )

= 1/3 . ( 1/2 - 1/52 )

= 1/3 . 25/52

= 25/156

\(\frac{1}{2.5}+\frac{1}{5.8}+....+\frac{1}{49.52}\)

\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{49.52}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{52}\right)=\frac{1}{3}.\left(\frac{26}{52}-\frac{1}{52}\right)=\frac{1}{3}.\frac{25}{52}=\frac{25}{156}\)

1320 ta

C = 6/2.5 + 6/5.8 + 6/8.11 +...+ 6/29.32
C = 2.(3/2.5 + 3/5.8 + 3/8.11 + ... + 3/29.32)
C = 2.(1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/29 - 1/32)
C = 2.(1/2 - 1/32)
C = 2.15/32
C = 15/16

Con cặc