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

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

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

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

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

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+....+\frac{1}{197.200}\)

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

\(=\frac{1}{5}-\frac{1}{200}\)

\(=\frac{40}{200}-\frac{1}{200}\)

\(=\frac{39}{200}\)

#H

Trả lời:

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{197.200}\)

\(=\frac{1.3}{5.8.3}+\frac{1.3}{8.11.3}+\frac{1.3}{11.14.3}+...+\frac{1.3}{197.200.3}\)

\(=\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{197.200}\right)\)

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

\(=\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{200}\right)=\frac{1}{3}.\frac{39}{200}=\frac{13}{200}\)

a) = ( 9/10 - 4/5) + ( 3/4 + 1/2)

= ( 9/10 - 8/10) + ( 3/4 + 2/4)

= 1/10 + 5/4 

= 2/20 + 25/20 = 27/20

b) = (4/5 + 4/15) +  (5/6 - 1/2)

= (12/15 - 4/15) + (5/6 - 3/6)

= 8/15 + 2/6 

= 16/30 + 10/30 

= 26/30 = 13/15

c) = 55 - 39 - 1 + 60 /75 = 75/75 = 1

d) = 47 + 35 - 36 + 15/42 = 61/42

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

3 x \(\frac{4}{11}\) = \(\frac{3}{1}\) x \(\frac{4}{11}\) = \(\frac{12}{11}\)

1 : \(\frac{5}{4}\) = \(\frac{1\times4}{5}\) = \(\frac{4}{5}\)

\(\frac{4}{5}\) x \(\frac{6}{7}\) + \(\frac{2}{3}\) = \(\frac{24}{35}\) + \(\frac{2}{3}\) = \(\frac{142}{105}\)

\(\frac{3}{4}\) + \(\frac{1}{2}\) : \(\frac{5}{6}\) = \(\frac{3}{4}\) + \(\frac{1}{2}\) x \(\frac{6}{5}\) = \(\frac{3}{4}\)+ \(\frac{6}{10}\) \(\frac{27}{20}\)

\(\frac{3}{4}\)+ \(\frac{1}{2}\) = \(\frac{3}{4}\) + \(\frac{2}{4}\)= \(\frac{5}{4}\)