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Chẳng ai quan tâm tới câu hỏi của tui. Buồn quá. Buồn không còn gì để tả. À mà có văn đâu mà tả? :))))
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{19}-\dfrac{1}{20}=\dfrac{19}{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=>M=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{2009\cdot2010}\)
`M=1/2-1/3+1/3-1/4+1/4-1/5+...+1/2009-1/2010`
`M=1/2-1/2010`
`M=502/1005`
![](https://rs.olm.vn/images/avt/0.png?1311)
B = 1/6 + 1/12 + 1/20 + ... + 1/90
B = 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/9.10
B = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/9 - 1/10
B = 1/2 - 1/10
B = 5/10 - 1/10
B = 4/10 = 2/5
Ủng hộ mk nha ♡_♡☆_☆
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![](https://rs.olm.vn/images/avt/0.png?1311)
ta có:
A= 1/6+1/12+1/20+1/30+1/42+1/56
= 1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8
= 1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8
= 1/2-1/8
= 3/8
vậy A= 3/8
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\)
\(\Rightarrow B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1-\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{100}\)
\(B=1-\frac{1}{100}=\frac{99}{100}\)
~ Hok tốt ~
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(A=\dfrac{-1}{3}+1+\dfrac{1}{3}=1\)
\(B=\dfrac{2}{15}+\dfrac{5}{9}-\dfrac{6}{9}=\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{18-15}{135}=\dfrac{3}{135}=\dfrac{1}{45}\)
\(C=\dfrac{-1}{5}+\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)
Bài 2:
a: \(=\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{2}{21}-\dfrac{10}{21}+\dfrac{3}{20}\)
\(=\left(\dfrac{1}{5}+\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{2}{21}-\dfrac{10}{21}\right)+\left(\dfrac{1}{2}+\dfrac{3}{20}\right)\)
\(=\dfrac{-8}{21}+\dfrac{13}{20}=\dfrac{113}{420}\)
b: \(B=\dfrac{21}{23}-\dfrac{21}{23}+\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)
Bài 3:
\(\dfrac{7}{3}-\dfrac{1}{2}-\left(-\dfrac{3}{70}\right)=\dfrac{7}{3}-\dfrac{1}{2}+\dfrac{3}{70}=\dfrac{490}{210}-\dfrac{105}{210}+\dfrac{9}{210}=\dfrac{394}{210}=\dfrac{197}{105}\)
\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}=\dfrac{5}{12}+\dfrac{3}{16}+\dfrac{3}{4}=\dfrac{20}{48}+\dfrac{9}{48}+\dfrac{36}{48}=\dfrac{65}{48}\)
Bài 4:
\(\dfrac{3}{4}-x=1\)
\(\Rightarrow-x=1-\dfrac{3}{4}\)
\(\Rightarrow x=-\dfrac{1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
\(x+4=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{5}-4\)
\(\Rightarrow x=-\dfrac{19}{5}\)
Vậy: \(x=-\dfrac{19}{5}\)
\(x-\dfrac{1}{5}=2\)
\(\Rightarrow x=2+\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{11}{5}\)
Vậy: \(x=\dfrac{11}{5}\)
\(x+\dfrac{5}{3}=\dfrac{1}{81}\)
\(\Rightarrow x=\dfrac{1}{81}-\dfrac{5}{3}\)
\(\Rightarrow x=-\dfrac{134}{81}\)
Vậy: \(x=-\dfrac{134}{81}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(B=\frac{1}{2}-\frac{1}{10}\)
\(B=\frac{2}{5}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{380}\)
\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
TL
\(\frac{1}{6}\)+\(\frac{1}{12}\)+\(\frac{1}{20}\)+....+\(\frac{1}{380}\)
=>\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+...+\(\frac{1}{19.20}\)
=>\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)+....+\(\frac{1}{19}\)-\(\frac{1}{20}\)
=>\(\frac{1}{2}\)-\(\frac{1}{20}\)
=>\(\frac{9}{20}\)
HT