\(S=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{29.32}\)

\(S=2.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\right)\)

\(S=2.\left(\frac{1}{2}-\frac{1}{32}\right)\)

\(S=2.\frac{15}{31}\Rightarrow S=\frac{15}{16}< 1\)

\(S=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{29.32}\)  

\(S=\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+...+\left(\frac{1}{29}-\frac{1}{32}\right)\)

\(S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\)

\(S=\frac{1}{2}-\frac{1}{32}\)

\(S=\frac{17}{32}< 1\)

S=6/2.5+6/5.8+6/8.11+...+6/29.32
S=2.(1/2-1/5+1/5-1/8+1/8-1/11+...+1/29-1/32)
S=2.(1/2+1/32)
S=2.15/32
S=30/32=15/16
=>S<1

=> 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:

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

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

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

\(S=\frac{6}{2.5}+\frac{6}{5.8}+.......+\frac{6}{29.32}\)

\(S=2\left(\frac{3}{2.5}+\frac{3}{5.8}+......+\frac{3}{29.32}\right)\)

\(S=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+......+\frac{1}{29}-\frac{1}{32}\right)\)

\(S=2\left(\frac{1}{2}-\frac{1}{32}\right)\)

\(S=2.\frac{15}{32}\)

\(S=\frac{15}{16}< 1\RightarrowĐPCM\)

Vậy \(S=\frac{15}{16}\)

\(S=2.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{29.32}\right)\)

\(S=2.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\right)\)

\(S=2.\left(\frac{1}{2}-\frac{1}{32}\right)\)

\(S=1-\frac{1}{16}< 1\)

Vậy \(S< 1\)

Bài 1 :

S = \(\frac{6}{2.5}+\frac{6}{5.8}+...+\frac{6}{29.32}\)

   = 2 . \(\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{29.32}\right)\)

   = 2 . \(\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\right)\)

   = 2 . \(\left(\frac{1}{2}-\frac{1}{32}\right)\)= ....

cả 2 cái cộng lại hay là từng cái một vậy bạn?

a) Ý bạn là: \(S_1=\frac{3}{4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)đúng không?

\(S_1=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

\(S_1=1-\frac{1}{43}< 1\left(đpcm\right)\)

b) \(S_2=\frac{6}{2\cdot5}+\frac{6}{5.8}+\frac{6}{8\cdot11}+...+\frac{6}{29\cdot32}\)

=>\(\frac{S_2}{2}=\frac{3}{2\cdot5}+\frac{3}{5.8}+\frac{3}{8\cdot11}+...+\frac{3}{29\cdot32}\)

\(\frac{S_2}{2}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\)

\(\frac{S_2}{2}=\frac{1}{2}-\frac{1}{32}=\frac{16}{32}-\frac{1}{32}=\frac{15}{32}\)

=>\(S_2=\frac{15}{32}\cdot2=\frac{15}{16}< 1\left(đpcm\right)\)

= 6/2 - 6/5 + 6/5 - 6/8 + ... + 6/62 - 6/65

= 6/2 - 6/65 = 189/65

6/2 - 6/5 + 6/5 - 6/8 + 6/8 - 6/11 + .... + 6/59 - 6/62 + 6/62 - 6/65

= 6/2 - 6/65

= 189/65

1/5.8+1/8.11+....+1/x(x+3)=1/6

=> 3/5.8 + 3/8.11 +.....+ 3/x(x+3)=1/2

=> 1/5-1/x+3=1/2

=> 1/x+3=1/5-1/2=-3/10=1/-10/3

=>x+3=-10/3=>x=-19/3

                   \(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)

                 \(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\cdot\left(x+3\right)}=\frac{1}{2}\)

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

                                                              \(\frac{1}{5}-\frac{1}{x+3}=\frac{1}{2}\)

                                                                          \(\frac{1}{x+3}=\frac{-3}{10}\)

\(\Rightarrow\left(-3\right)\left(x+3\right)=10\)

\(\Rightarrow x+3=\frac{-10}{3}\)

\(\Rightarrow x=\frac{-19}{3}\)

Vậy \(x=\frac{-19}{3}\)