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\(A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{98}\)
\(A=\dfrac{1}{2}-\dfrac{1}{98}=\dfrac{49}{98}-\dfrac{1}{98}=\dfrac{48}{98}=\dfrac{24}{49}\)
\(A=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{92\cdot95}+\dfrac{3}{95\cdot98}\right)\\ A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{95}-\dfrac{1}{98}\right)\\ A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{98}\right)=\dfrac{1}{3}\cdot\dfrac{24}{49}=\dfrac{8}{49}\)
\(A=\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{92\cdot95}+\dfrac{1}{95\cdot98}\)
\(A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{2}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{92}+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{98}\)
\(A=\dfrac{1}{2}-\dfrac{1}{98}\)
\(A=\dfrac{49}{98}-\dfrac{1}{98}\)
\(A=\dfrac{48}{98}\)
\(A=\dfrac{24}{49}\)
Giải thích các bước giải:
A =1/2.5 + 1/5.8 + 1/8.11 + … +1/92.95 + 1/95.98
=1/3 . (1/2-1/5+1/5-1/8+1/8-1/11+…+1/92-1/95+1/95-1/98)
=1/3 . (1/2 – 1/98 )
=1/3 . 24/49
=8/49`
vậy `A=8/49`
\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}\)
= \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}\)
\(=\dfrac{1}{2}-\dfrac{1}{17}\)
\(=\dfrac{15}{34}\)
Vì \(\dfrac{15}{34}< \dfrac{1}{2}=>\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot27}< \dfrac{1}{2}\)
a: \(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{201}-\dfrac{1}{203}=\dfrac{202}{203}\)
b: \(=-4\left(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{2015\cdot2018}\right)\)
\(=-\dfrac{4}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{2015\cdot2018}\right)\)
\(=\dfrac{-4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{2015}-\dfrac{1}{2018}\right)\)
\(=\dfrac{-4}{3}\cdot\dfrac{504}{1009}=-\dfrac{672}{1009}\)
\(x\) \((\)\(\dfrac{3}{2.5}\) \(+
\) \(\dfrac{3}{5.8}\) \(+\) \(\dfrac{3}{8.11}\) \(+\) \(\dfrac{3}{11.14}\)\()\) \(=\) \(\dfrac{1}{21}\)
\(x\) \((\)\(\dfrac{1}{2}\) \(-\) \(\dfrac{1}{5}\) \(+\) \(\dfrac{1}{5}\) \(-\) \(\dfrac{1}{8}\) \(+\) \(\dfrac{1}{8}\) \(-\) \(\dfrac{1}{11}\) \(+\) \(\dfrac{1}{11}\) \(-\) \(\dfrac{1}{14}\)\()\) \(=\) \(\dfrac{1}{21}\)
\(x\) \((\)\(\dfrac{1}{2}\) \(-\) \(\dfrac{1}{14}\)\()\) \(=\) \(\dfrac{1}{21}\)
\(x\) x \(\dfrac{3}{7}\) \(=\) \(\dfrac{1}{21}\)
\(x\) \(=\) \(\dfrac{1}{21}\) \(:\) \(\dfrac{3}{7}\)
\(x\) \(=\) \(\dfrac{1}{9}\)
\(P=\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{97\cdot100}\)
\(P=\left(\dfrac{1}{2}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}-\dfrac{1}{8}\right)+...+\left(\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(P=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{97}-\dfrac{1}{100}\)
\(P=\dfrac{1}{2}-\dfrac{1}{100}\)
\(P=\dfrac{50}{100}-\dfrac{1}{100}\)
\(P=\dfrac{49}{100}\)
\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{3n+2}{6n+4}-\dfrac{2}{6n+4}\right)\)
\(=\dfrac{1}{3}.\dfrac{3n}{6n+4}\)
\(=\dfrac{n}{6n+4}\) ( đpcm )
Vậy...
\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{y\left(y+3\right)}=\dfrac{98}{1545}\)
\(\Leftrightarrow\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{y\left(y+3\right)}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{y}-\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{y+3}{5\left(y+3\right)}-\dfrac{5}{5\left(y+3\right)}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{y+3-5}{5\left(y+3\right)}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{y-2}{5\left(y+3\right)}=\dfrac{98}{515}\)
\(\Leftrightarrow515\left(y-2\right)=98.5\left(y+3\right)\)
\(\Leftrightarrow515y-1030=490y+1470\)
\(\Leftrightarrow25y-2500=0\\ \Leftrightarrow25y=2500\\ \Leftrightarrow y=100\)
\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{y+3}\right)=\dfrac{98}{1545}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{y+3}=\dfrac{98}{515}\)
\(\Leftrightarrow\dfrac{1}{y+3}=\dfrac{1}{103}\)
hay x=100
3A=3/2.5+...+3/2018.2021
3A=1/2-1/5+1/5-...+1/2018-1/2021
3A=1/2-1/2021 sau tự tính A
3A= 1/2- 1/5 + 1/5- 1/8+ 1/8 -1/11+...+ 1/2012- 1/2015 +1/2015- 1/2018-1/2021
3A =1/2 -1/2021
3A = 2019/ 4042
=> 2019/4042 : 3 = 673/4042
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