Lời giải:

$A=\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+....+\frac{1}{197.200}$

$3A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+....+\frac{3}{197.200}$

$3A=\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+...+\frac{200-197}{197.200}$

$=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}$

$=\frac{1}{5}-\frac{1}{200}=\frac{39}{200}$

$A=\frac{13}{200}$

Ta có: \(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{197\cdot200}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{197\cdot200}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{200}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{39}{200}=\dfrac{13}{200}\)

\(\frac{1}{5\times8}+\frac{1}{8\times11}+\frac{1}{11\times14}+...+\frac{1}{197\times200}\)

\(=\frac{1}{3}\times\left(\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+...+\frac{3}{197\times200}\right)\)

\(=\frac{1}{3}\times\left(\frac{8-5}{5\times8}+\frac{11-8}{8\times11}+\frac{14-11}{11\times14}+...+\frac{200-197}{197\times200}\right)\)

\(=\frac{1}{3}\times\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(=\frac{1}{3}\times\left(\frac{1}{5}-\frac{1}{200}\right)\)

\(=\frac{13}{200}\)

x.\(\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\right)=-1\frac{3}{5}\)

x.\(\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+\frac{17-14}{14.17}\right)=\frac{-8}{5}\)

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

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

x.\(\left(\frac{17}{34}-\frac{2}{34}\right)=\frac{-8}{5}\)

x.\(\frac{15}{34}=\frac{-8}{5}\)

x\(=\frac{-8}{5}:\frac{15}{34}\)

x\(=\frac{-8}{5}.\frac{34}{15}\)

x\(=\frac{-272}{75}\)

Vậy x\(=\frac{-272}{75}\)

27,3 cộng 5,7 = bao nhiêu

fffff

đúng rồi đó

Đúng rồi người gì đâu mà vô đạo đức

a) Đặt A= \(\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{36}\)

\(\dfrac{1}{2}\)A=\(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\)

\(\dfrac{1}{2}\)A=\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)

\(\dfrac{1}{2}\)A=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

\(\dfrac{1}{2}\)A=\(\dfrac{1}{2}-\dfrac{1}{9}\)

\(\dfrac{1}{2}\)A=\(\dfrac{7}{18}\)

A=\(\dfrac{7}{9}\)

B=\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{20}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{19}{20}\)

\(=\dfrac{1.2.3....19}{2.3.4.....20}\)

\(=\dfrac{1.2.3....19:\left(2.3.....19\right)}{2.3.4.....20:\left(2.3.4.....19\right)}\)

\(=\dfrac{1}{20}\)