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

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

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

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

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

\(S=\frac{15}{16}< 1\RightarrowĐPCM\)

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

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

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

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

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

S=6/2*5+6/5*8+...+6/29*32

=6/3*(3/2*5+3/5*8+...+3/29*32)

=2*(1/2-1/5+1/5-1/8+...+1/29-1/32)

=2*(1/2-1/32)=2*15/32

=15/16<1

S=6/2*5+6/5*8+...+6/29*32,c

=6/3*(3/2*5+3/5*8+...+3/29*32)

=2*(1/2-1/5+1/5-1/8+...+1/29-1/32)

=2*(1/2-1/32)

=2*15/32

=15/16<1

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

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

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

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

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

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

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

Ta có: 9/2.5  + 9/5.8 + 9/9.11+...+ 9/29.32

      =>9.(1/2 -1/5 + 1/5 - 1/8 +...+1/29 -1/32)

      =>9.( 1/2 -1/32)

      =>9. 15/32

      =>45/32 

Suy ra 45/32 >1 nên S>1

Ta có: \(S=\dfrac{9}{2\cdot5}+\dfrac{9}{5\cdot8}+\dfrac{9}{8\cdot11}+...+\dfrac{9}{29\cdot32}\)

\(=3\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{29\cdot32}\right)\)

\(=3\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{29}-\dfrac{1}{32}\right)\)

\(=3\left(\dfrac{1}{2}-\dfrac{1}{32}\right)\)

\(=3\cdot\dfrac{15}{32}=\dfrac{45}{32}>1\)

1/2.5+1/5.8+1/8.11+...+1/29.32 (khoảng cách từ 2-5;5-8;8-11;...;29-32 là 3) suy ra

=1/3(1/2-1/5+1/5-1/8+1/8-1/11+...+1/29-1/32)  (-1/5+1/5;-1/8+1/8;-1/11+1/11=0) suy ra =1/3(1/2-1/32)=1/3.15/32=5/32

1/2.5+1/5.8+1/8.9+............+1/29.32

=1/2-1/5+1/5-1/8+...............+1/29-1/32

=1/2-1/32

=15/32

ai tích mk=>mk tích lại

\(\frac{12}{2.5}+\frac{12}{5.8}+\frac{12}{8.11}+...+\frac{12}{29.32}\)

\(=4.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{29.32}\right)\)

\(=4.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\right)\)

\(=4.\left(\frac{1}{2}-\frac{1}{32}\right)\)

\(=4.\frac{15}{32}\)

\(=\frac{15}{8}\)

_Chúc bạn học tốt_

\(\frac{12}{2.5}+\frac{12}{5.8}+\frac{12}{8.11}+....+\frac{12}{29.32}\)

\(=4\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{29.32}\right)\)

\(=4\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\right)\)

\(=4\left(\frac{1}{2}-\frac{1}{32}\right)\)

\(=4.\frac{15}{32}=\frac{15}{8}\)

Đặt A= 12/2*5+12/5*8+12/8*11+....+12/29*32

       A= 4*(3/2*5+3/5*8+3/8*11+...+3/29*32)

       A = 4*(1/2-1/5+1/5-1/8+1/8-1/11+....+1/29-1/32)

       A= 4*(1/2-1/32)

       A= 4*15/32

       A= 15/8