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Ta có công thức tổng quát:
\(\dfrac{k}{n\cdot\left(n+k\right)}=\dfrac{1}{n}-\dfrac{1}{n+k}\)
\(a,A=\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}\\ =\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)\\ =\dfrac{1}{3}\cdot\dfrac{x-2}{5\left(x+3\right)}\\ =\dfrac{x-2}{15\left(x+3\right)}\)
Theo đề bài ta có:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
=13/12x14/13x15/14x16/15x...x2006/2005x2007/2006x2008/2007
=2008/12
=502/3
A = 1\(\dfrac{1}{12}\) \(\times\) 1\(\dfrac{1}{13}\) \(\times\) 1\(\dfrac{1}{14}\) \(\times\) 1\(\dfrac{1}{15}\) \(\times\) ... \(\times\) 1\(\dfrac{1}{2005}\) \(\times\) 1\(\dfrac{1}{2006}\) \(\times\) 1\(\dfrac{1}{2007}\)
A = ( 1 + \(\dfrac{1}{12}\)) \(\times\) ( 1 + \(\dfrac{1}{13}\)) \(\times\) ( 1 + \(\dfrac{1}{14}\)) \(\times\)...\(\times\) ( 1 + \(\dfrac{1}{2006}\))\(\times\)(1+\(\dfrac{1}{2007}\))
A = \(\dfrac{13}{12}\) \(\times\) \(\dfrac{14}{13}\) \(\times\) \(\dfrac{15}{14}\) \(\times\) ...\(\times\) \(\dfrac{2007}{2006}\) \(\times\) \(\dfrac{2008}{2007}\)
A = \(\dfrac{13\times14\times15\times...\times2007}{13\times14\times15\times...\times2007}\) \(\times\) \(\dfrac{2008}{12}\)
A = 1 \(\times\) \(\dfrac{502}{3}\)
A = \(\dfrac{502}{3}\)
TL:
\(\frac{12}{100}\)= 0,12
\(\frac{5}{100}\)= 0,05
\(\frac{306}{1000}\)= 0,306
-HT-
Do a < b nên \(\frac{3}{a}>\frac{3}{b}\) hay \(\frac{3}{b}< \frac{3}{a}\).
Ta thấy \(\frac{39}{40}=\frac{3}{a}+\frac{3}{b}< \frac{3}{a}+\frac{3}{a}=\frac{6}{a}\) nên suy ra \(\frac{39}{40}< \frac{6}{a}\Rightarrow\frac{78}{80}< \frac{78}{13a}\Rightarrow80>13a\)
Mà \(\frac{3}{a}< \frac{39}{40}\Rightarrow\frac{39}{13a}< \frac{39}{40}\Rightarrow13a>40\)
Nên 80 > 13a > 40. Vậy a = { 4 ; 5 ; 6 }
- Với a = 4 thì \(b=\frac{3}{\frac{39}{40}-\frac{3}{4}}=\frac{3}{\frac{9}{40}}=\frac{120}{9}=\frac{40}{3}\) ( Loại vì không phải số tự nhiên )
- Với a = 5 thì \(b=\frac{3}{\frac{39}{40}-\frac{3}{5}}=\frac{3}{\frac{3}{8}}=\frac{24}{3}=8\) ( Hợp lệ )
- Với a = 6 thì \(b=\frac{3}{\frac{39}{40}-\frac{3}{6}}=\frac{3}{\frac{19}{40}}=\frac{120}{19}\) ( Loại vì không phải số tự nhiên )
Vậy a = 5 ; b = 8
\(\overline{36,75x4}\) < \(\dfrac{367544}{10000}\)
\(\overline{36,75x4}\) < 36,7544
vì 36,7504 < 36,7514< 36,7524< 36,7534< 36,7544
\(x\) = 0; 1; 2; 3
b, \(\overline{ab5,728}\) < \(\overline{ab5,7x4}\) < \(\overline{ab5,755}\)
vì \(\overline{ab5,728}\) < \(\overline{ab5,734}\) < \(\overline{ab5,744}\) < \(\overline{ab5,754}\) < \(\overline{ab5,755}\)
vậy \(x\) = 3; 4; 5