A=...

<=>\(A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{1}{17.20}\right)\)

<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)

<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)

<=>\(A=\frac{1}{6}-\frac{1}{60}< \frac{1}{6}< 1\)

sai ùi 

\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)

\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(A=\frac{1}{3}.\frac{9}{20}\)

\(A=\frac{3}{20}\)

Vì \(\frac{3}{20}< 1\Rightarrow A< 1\)

- A ở trên giữa các phân số là dấu " + " nha mấy bạn !

\(7\frac{x}{2.5}+7\frac{x}{5.8}+.....+7.\frac{x}{17.20}=\frac{21}{10}\)

\(7\left(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}\right)=\frac{21}{10}\)

\(\frac{x}{2.5}+\frac{x}{5.8}+...+\frac{x}{17.20}=\frac{21}{70}\)

\(\frac{x.3}{2.5.3}+\frac{x.3}{5.8.3}+...+\frac{x.3}{17.20.3}=\frac{21}{70}\)

\(x.\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\right)=\frac{21}{70}\)

\(x.\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{21}{70}\)

\(x.\frac{1}{3}.\frac{9}{20}=\frac{21}{70}\)

=> \(x=2\)

\(x=\frac{7x}{2}\)\(-\frac{7x}{5}+\)\(\frac{7x}{5}\)\(-\frac{7x}{8}\)\(+\frac{7x}{8}\)\(-\frac{7x}{11}\)\(+\frac{7x}{11}\)\(-\frac{7x}{14}\)\(+\frac{7x}{14}\)\(-\frac{7x}{17}+\)\(\frac{7x}{17}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x}{2}\)\(-\frac{7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x.10}{20}\)\(+\frac{7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x.10+7x}{20}\)\(=\frac{21}{10}\)

\(x=\frac{7x.\left(10+2\right)}{20.2}\)\(=\frac{7x.12}{40}\)\(=\frac{21}{10}\)

\(=>\frac{7x.12:4}{40:4}=\)\(\frac{21}{10}\)

\(=>x=1\)

Đặt \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\)

\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)

\(A=\frac{1}{2}-\frac{1}{17}\)

\(A=\frac{15}{34}\)

= \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)= \(\frac{1}{2}-\frac{1}{17}\)=\(\frac{15}{34}\)

Ta có: \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{14}-\frac{1}{17}\)

\(=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)\(< \frac{17}{34}=\frac{1}{2}\)

\(\Rightarrow\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{14.17}< \frac{1}{2}\)

Vậy:..........................................(đpcm)

Đề bài là tìm x nhé các bạn, các bạn giải giúp mik nhé!!

Bài làm

\(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}=\frac{1}{21}\)

\(\Leftrightarrow\frac{x}{2}-\frac{x}{5}+\frac{x}{5}-\frac{x}{8}+\frac{x}{8}-\frac{x}{11}+\frac{x}{11}-\frac{x}{14}=\frac{1}{21}\)

\(\Leftrightarrow\frac{x}{2}-\frac{x}{14}=\frac{1}{21}\)

\(\Leftrightarrow\frac{7x}{14}-\frac{x}{14}=\frac{1}{21}\)

\(\Leftrightarrow\frac{6x}{14}=\frac{1}{21}\)

\(\Leftrightarrow126x=14\)

\(\Leftrightarrow x=\frac{1}{9}\)

Học tôt 

Bài làm:
Ta có: \(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}=\frac{1}{21}\)

\(\Leftrightarrow x\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}\right)=\frac{1}{21}\)

\(\Leftrightarrow x\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)

\(\Leftrightarrow x\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)

\(\Leftrightarrow\frac{3}{7}x=\frac{1}{21}\)

\(\Leftrightarrow x=\frac{1}{9}\)

Học tốt!!!!
 

xét vế trái

ta có:đề\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{14}-\frac{1}{17}\) 

\(=\frac{1}{2}-\frac{1}{17}< < \frac{1}{2}\)

vậy vế trái bé hơn \(\frac{1}{2}\)

P/S:  \(< < \)là luôn luôn bé hơn nha

k mình nha bạn

Thiengl2015#

Ta có :

\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{14}-\frac{1}{17}\)

\(=\frac{1}{2}-\frac{1}{17}\)

Mà \(\frac{1}{2}-\frac{1}{17}< \frac{1}{2}\)

Nên \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}< \frac{1}{2}\left(đpcm\right)\)

\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{14\cdot17}\)

=\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{14}-\frac{1}{17}\)

=\(\frac{1}{2}-\frac{1}{17}\)

=\(\frac{17}{34}-\frac{2}{34}\)

=\(\frac{15}{34}\)

Mà \(\frac{1}{2}=\frac{17}{34}>\frac{15}{34}\)

\(\Rightarrowđpcm\)

S = 1/3 . (1/2 - 1/5 + 1/5 - 1/8 + ... + 1/17 - 1/20)

   = 1/3 . (1/2 - 1/20)

   = 1/3 . 9/20

   = 3/20

\(3S=\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{20-17}{17.20}\)

\(3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)

\(S=\frac{9}{20}:3=\frac{3}{20}\)

\(S=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...........+\frac{3}{17.20}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.........+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(=\frac{1}{3}.\left(\frac{10}{20}-\frac{1}{20}\right)\)

\(=\frac{1}{3}.\frac{9}{20}\)

\(-\frac{3}{20}\)