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\(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{37.40}< \dfrac{1}{5}\)
=\(\dfrac{3}{3}\left(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{37.40}\right)\)
=\(\dfrac{1}{3}\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{37.40}\right)\)
=\(\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)\)
=\(\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{40}\right)\)
=\(\dfrac{3}{40}< \dfrac{1}{3}\)
1.
E = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{3}{7.10}\) + \(\dfrac{3}{10.13}\) + \(\dfrac{3}{13.16}\) + \(\dfrac{3}{16.19}\) + \(\dfrac{3}{19.22}\)
E = 1 - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{10}\) + ... +\(\dfrac{1}{19}\) - \(\dfrac{1}{22}\)
E = 1 - \(\dfrac{1}{22}\)
E = \(\dfrac{21}{22}\)
2.
(x - 4)(x - 5) = 0
TH1:
x - 4 = 0 => x = 4
TH2:
x - 5 = 0 => x = 5
Vậy: x = 4 hoặc x = 5
Ta có : A = 2/ 4.7 + 2/ 7.10 + ... + 2/ 73.76 .
⇒ 3/2 A = 3/2 . ( 2/ 4.7 + 2/ 7.10 + ... + 2/ 73.76 ) .
⇒ 3/2 A = 3/ 4.7 + 3/ 7.10 + ... + 3/ 73.76 .
= 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/ 73 - 1/76 .
= 1/4 - 1/76 .
= 19/76 - 1/76 .
= 9/38 .
Do đó : A = 9/38 : 3/2 .
= 9/38 . 2/3 .
= 3/19 .
Vậy A = 3/19 .
\(A=3.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)
\(A=3.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(A=3.\left(1-\dfrac{1}{100}\right)\)
\(A=3.\dfrac{99}{100}=\dfrac{297}{100}\)
\(A=\dfrac{1}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{25\cdot28}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{6}{28}=\dfrac{2}{28}=\dfrac{1}{14}\)
`3A = 3/(4.7) + 3/(7.10) + .. + 3/(25.28)`
`3A = 1/4 - 1/7 + 1/7 - 1/10 +... + 1/25 - 1/28`
`3A = 3/14`
`A = 1/14.`
Bài 1:
Ta có: \(A=\dfrac{2011+2012}{2012+2013}=\dfrac{2011}{2012+2013}+\dfrac{2012}{2012+2013}\)
Dễ thấy:
\(\dfrac{2011}{2012+2013}< \dfrac{2011}{2012};\dfrac{2012}{2012+2013}< \dfrac{2012}{2013}\)
\(\Rightarrow A=\dfrac{2011}{2012+2013}+\dfrac{2012}{2012+2013}< B=\dfrac{2011}{2012}+\dfrac{2012}{2013}\)
Bài 2:
\(S=\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{37\cdot40}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{37\cdot40}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{40}\right)=\dfrac{1}{3}\cdot\dfrac{9}{40}=\dfrac{3}{40}< \dfrac{1}{3}\)
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...................+\dfrac{3}{n\left(n+1\right)}\)
\(\Rightarrow S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.............+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(\Rightarrow S=1-\dfrac{1}{n+1}< 1\)
\(\Rightarrow S< 1\rightarrowđpcm\)
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{n.\left(n+1\right)}\)
\(S=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...-\dfrac{1}{n+1}\)
\(S=1-\dfrac{1}{n+1}\)\(< 1\)
\(\Leftrightarrow S< 1\)
tik cho mik nhé
Ta có:
\(S=\dfrac{3}{3}.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{43}-\dfrac{1}{46}\right)\)
\(S=1.\left(\dfrac{1}{1}-\dfrac{1}{46}\right)\)
\(S=1.\dfrac{45}{46}=\dfrac{45}{46}\)
Vì \(\dfrac{45}{46}< \dfrac{46}{46}\) nên \(\dfrac{45}{46}< 1\).
Vậy S < 1.
\(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{43.46}\)
\(S=\dfrac{3}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{43.46}\right)\)
Ta thấy:
\(\dfrac{3}{1.4}=1-\dfrac{1}{4};\dfrac{3}{4.7}=\dfrac{1}{4}-\dfrac{1}{7};\dfrac{3}{7.10}=\dfrac{1}{7}-\dfrac{1}{10};\)
\(...;\dfrac{3}{43.46}=\dfrac{1}{43}-\dfrac{1}{46}\)
\(\Rightarrow S=1\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{43}-\dfrac{1}{46}\right)\)
\(\Rightarrow S=1\left(1-\dfrac{1}{46}\right)\)
\(\Rightarrow S=1.\dfrac{45}{46}=\dfrac{45}{46}\)
\(A=\dfrac{2}{1.4}+\dfrac{2}{4.7}+\dfrac{2}{7.10}+...+\dfrac{2}{\left(3x+1\right).\left(3x+4\right)}\)=\(\dfrac{1344}{2017}\)
\(A=\dfrac{2}{3}(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{3x+1}-\dfrac{1}{3x+4}\))=\(\dfrac{1344}{2017}\)
\(A=\dfrac{2}{3}(1-\dfrac{1}{3x+4})\)=\(\dfrac{1344}{2017}\)
\(A=1-\dfrac{1}{3x+4}=\dfrac{1344}{2017}:\dfrac{2}{3}\)
\(A=1-\dfrac{1}{3x+4}=\dfrac{2016}{2017}\)
\(A=\dfrac{1}{3x+4}=1-\dfrac{2016}{2017}\)
\(A=\dfrac{1}{3x+4}=\dfrac{1}{2017}\)
\(\Rightarrow\)\(3x+4=2017\)
\(3x=2017-4\)
\(3x=2013\)
\(x=671\)
\(\Leftrightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\)
\(\rightarrowđpcm\)
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\(\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{25.28}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{25.28}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{3}{14}\)
\(=\dfrac{1}{14}\)
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