![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{12}{2.5}+\dfrac{12}{5.8}+.......+\dfrac{12}{65.68}\)
\(=4\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+......+\dfrac{3}{65.68}\right)\)
\(=4\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+.......+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(=4\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)
\(=4.\dfrac{33}{68}=\dfrac{33}{17}\)
Dễ quá! Vì mình là một CTV bên Học toán với OnlineMath nên bài này easy!! :")))
\(\dfrac{12}{2.5}+\dfrac{12}{5.8}+\dfrac{12}{8.11}+...+\dfrac{12}{65.68}\)
\(\Leftrightarrow4\left(\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{65.68}\right)\)
\(\Leftrightarrow4\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(=4\left(1-\dfrac{1}{68}\right)=4.\dfrac{67}{68}=\dfrac{67}{17}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\frac{12}{2.5}+\frac{12}{5.8}+\frac{12}{8.11}+...+\frac{12}{65.68}\)
\(=\)\(4\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{65.68}\right)\)
\(=\)\(4\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{65}-\frac{1}{68}\right)\)
\(=\)\(4\left(\frac{1}{2}-\frac{1}{68}\right)\)
\(=\)\(2-\frac{1}{17}\)
\(=\)\(\frac{35}{17}\)
Vậy \(\frac{12}{2.5}+\frac{12}{5.8}+\frac{12}{8.11}+...+\frac{12}{65.68}=\frac{35}{17}\)
Chúc bạn học tốt ~
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=2.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{29.32}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=1-\frac{1}{16}< 1\)
Vậy \(S< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{29.32}\)
\(S=2.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=2.\frac{15}{31}\Rightarrow S=\frac{15}{16}< 1\)
\(S=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{29.32}\)
\(S=\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+...+\left(\frac{1}{29}-\frac{1}{32}\right)\)
\(S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\)
\(S=\frac{1}{2}-\frac{1}{32}\)
\(S=\frac{17}{32}< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\frac{6}{2.5}+\frac{6}{5.8}+.......+\frac{6}{29.32}\)
\(S=2\left(\frac{3}{2.5}+\frac{3}{5.8}+......+\frac{3}{29.32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+......+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=2.\frac{15}{32}\)
\(S=\frac{15}{16}< 1\RightarrowĐPCM\)
Vậy \(S=\frac{15}{16}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
= 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + .... + 1/29 - 1/32
= 1/2 - 1/32
= ..... ( tự bấm máy tính nhé )
Ta có: \(E=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{29}-\frac{1}{32}\)
\(\Rightarrow E=\frac{1}{2}-\frac{1}{32}\)
\(\Rightarrow E=\frac{16-1}{32}=\frac{15}{32}\)
Vậy \(E=\frac{15}{32}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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) \(S_2=\frac{6}{2\cdot5}+\frac{6}{5.8}+\frac{6}{8\cdot11}+...+\frac{6}{29\cdot32}\)
=>\(\frac{S_2}{2}=\frac{3}{2\cdot5}+\frac{3}{5.8}+\frac{3}{8\cdot11}+...+\frac{3}{29\cdot32}\)
\(\frac{S_2}{2}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\)
\(\frac{S_2}{2}=\frac{1}{2}-\frac{1}{32}=\frac{16}{32}-\frac{1}{32}=\frac{15}{32}\)
=>\(S_2=\frac{15}{32}\cdot2=\frac{15}{16}< 1\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 . \(\left(\frac{1}{2}-\frac{1}{32}\right)\)= ....
\(\frac{12}{2.5}+\frac{12}{5.8}+\frac{12}{8.11}+...+\frac{12}{29.32}\)
\(=4.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{29.32}\right)\)
\(=4.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\right)\)
\(=4.\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(=4.\frac{15}{32}\)
\(=\frac{15}{8}\)
_Chúc bạn học tốt_
\(\frac{12}{2.5}+\frac{12}{5.8}+\frac{12}{8.11}+....+\frac{12}{29.32}\)
\(=4\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{29.32}\right)\)
\(=4\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\right)\)
\(=4\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(=4.\frac{15}{32}=\frac{15}{8}\)