d ( 1-1/2)x(1-1/3)x(1-1/4)x......x(1-1/2018)

 = 1/2x2/3x3/4x...x2017/2018

=\(\frac{1x2x3x....x2017}{2x3x4x....x2018}\)

=  \(\frac{1}{2018}\)

e , 1+4+7+...+100

= dãy có  số số hạng là 

 (100-1):3+1=34 ( số số hạng)

tổng là : (100+1 ) x 34 : 2 =1717

=>1717

bậy bạ

?

><

\(\dfrac{x}{2\times5}+\dfrac{x}{5\times8}+\dfrac{x}{8\times11}+\dfrac{x}{11\times14}+...+\dfrac{x}{32\times35}=\dfrac{33}{70}\)

\(\dfrac{x}{3}\cdot\left(\dfrac{3}{2\times5}+\dfrac{3}{5\times8}+\dfrac{3}{8\times11}+\dfrac{3}{11\times14}+...+\dfrac{3}{32\times35}\right)=\dfrac{33}{70}\)

\(\dfrac{x}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{32}-\dfrac{1}{35}\right)=\dfrac{33}{70}\)

\(\dfrac{x}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{35}\right)=\dfrac{33}{70}\)

\(\dfrac{x}{3}\cdot\dfrac{33}{70}=\dfrac{33}{70}\)

\(\dfrac{x}{3}=\dfrac{33}{70}:\dfrac{33}{70}\)

\(\dfrac{x}{3}=1\)

\(x=3\)

\(\Leftrightarrow x\cdot\dfrac{1}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{32\cdot35}\right)=\dfrac{33}{70}\)

=>\(x\cdot\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{32}-\dfrac{1}{35}\right)=\dfrac{33}{70}\)
=>\(x\cdot\dfrac{1}{3}\cdot\dfrac{33}{70}=\dfrac{33}{70}\)

=>x=3

\(\frac{3}{2\times5}+\frac{2}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{602\times605}\)

\(=\frac{5-2}{2\times5}+\frac{8-5}{5\times8}+\frac{11-8}{8\times11}+...+\frac{605-602}{602\times605}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\)

\(=\frac{1}{2}-\frac{1}{605}=\frac{603}{1210}\)

\(\frac{4}{3\times7}+\frac{5}{7\times12}+\frac{1}{12\times13}+\frac{2}{13\times15}\)

\(=\frac{7-4}{3\times7}+\frac{12-7}{7\times12}+\frac{13-12}{12\times13}+\frac{15-13}{13\times15}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)

\(=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)

Ta có: \(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)

\(\Leftrightarrow x\cdot\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right)=\frac{1}{21}\)

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

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

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

\(\Leftrightarrow x=\frac{1}{21}:\frac{3}{7}=\frac{1}{21}\cdot\frac{7}{3}=\frac{7}{63}=\frac{1}{9}\)

Vậy: \(x=\frac{1}{9}\)

\(B=\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{26\cdot29}\)

\(B=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{29}\)

\(B=\dfrac{1}{2}-\dfrac{1}{29}\)

\(B=\dfrac{27}{58}\)

B= 3/2x5 + 3/5x8+ 3/8x11 + ... + 3/26x29

B= 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/26 - 1/29

B= 1/2-1/29

B=27/58

\(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+....+\frac{1}{97\cdot100}\)

\(=\frac{5-2}{2\cdot5}+\frac{8-5}{5\cdot8}+\frac{11-8}{8\cdot11}+...+\frac{100-97}{97\cdot100}\)

\(=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=\frac{1}{3}\cdot\frac{49}{100}=\frac{49}{300}\)