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

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

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

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

\(B=17-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{53}\right)=17-\frac{1}{3}.\frac{51}{106}=17-\frac{17}{106}=17\left(1-\frac{1}{106}\right)=17.\frac{105}{106}=\frac{1785}{106}\)

B=9/2,5+39/5,8+87/8,11+...+2649/50,53

B=2,5-1/2,5+5,8-1/5,8+8,11-1/8,11+...+50,53-1/50,53

B=1-1/2,5+1-1/5,8+1-1/8,11+...+1-1/50,53

B=17-1/3(3/2,5+3/5,8+3/8,11+...+3/50,53

B=17-1/3(1/2-1/5+1/5-1/8+...+1/50-1/53)

B=17-1/3(1/2-1/53)

B=1785/106

Học tốt nha bạn ~!!!

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

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

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

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

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

nhân 3 vào cả hai vế 

\(D=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{1979.1982} \)

    \(=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{1979.1982}\right)\)

    \(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{1979}-\frac{1}{1982}\right)\)

    \(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{1982}\right)\)

    \(=\frac{165}{991}\)

A=(1/2-1/5+1/5-1/8+1/8-....+1/50-1/51)

  = 1/2-1/51

  = 51/102 - 2/102

  = 49/102

B=1.4/1.5.4+1.4/5.9.4+...+1.4/41.45.4

  = 1/4(1-1/5+1/5-1/9+1/9-...+1/41-1/45)

  = 1/4(1-1/45)

  = 1/4.44/45

  = 11/45

=> 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:

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

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

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

Đề hình như bị sai ban ơi sửa lại

\(A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{92.95}\)

\(A=3\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)

\(A=3.\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)

\(A=\dfrac{1}{2}-\dfrac{1}{95}\)

\(A=\dfrac{93}{190}\)

\(B=\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{92.95}\)

\(3B=2\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}\right)\)

\(3B=2.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\right)\)

\(3B=2\left(\dfrac{1}{2}-\dfrac{1}{95}\right)\)

\(3B=2.\dfrac{93}{190}\)

\(3B=\dfrac{93}{95}\)

\(\Rightarrow B=\dfrac{31}{95}\)

a) \(\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+.......+\frac{6}{44.47}+\frac{6}{47.50}\)

\(=2\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+......+\frac{3}{44.47}+\frac{3}{47.50}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{44}-\frac{1}{47}+\frac{1}{47}-\frac{1}{50}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{50}\right)\)

\(=1-\frac{1}{25}\)

\(=\frac{24}{25}\)

đặt \(A=\frac{1}{9.11}+\frac{1}{11.13}+........+\frac{1}{41.43}+\frac{1}{43.45}\)

\(2A=\frac{2}{9.11}+\frac{2}{11.13}+.......+\frac{2}{41.43}+\frac{2}{43.45}\)

\(2A=\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+......+\frac{1}{41}-\frac{1}{43}+\frac{1}{43}-\frac{1}{45}\)

\(2A=\frac{1}{9}-\frac{1}{45}\)

\(2A=\frac{4}{45}\)

\(A=\frac{4}{45}\div2\)

\(A=\frac{2}{45}\)

khó thế ko giải được

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