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TL:
\(\frac{12}{100}\)= 0,12
\(\frac{5}{100}\)= 0,05
\(\frac{306}{1000}\)= 0,306
-HT-
\(\frac{27}{31}=\frac{2727}{3131}=\frac{272727}{313131}\)
\(\frac{27}{31}=\frac{27.101}{31.101}=\frac{27.10101}{31.10101}\)
\(\frac{27}{31}=\frac{27}{31}=\frac{27}{31}\)
\(\Rightarrow\frac{27}{31}=\frac{2727}{3131}=\frac{272727}{313131}\)
\(\frac{27}{31}=\frac{2727}{3131}=\frac{272727}{313131}\)
\(\frac{27}{31}=\frac{27\cdot101}{31\cdot101}=\frac{27\cdot10101}{31\cdot10101}\)
\(\Rightarrow\frac{27}{31}=\frac{2727}{3131}=\frac{272727}{313131}\)
=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 = \(\dfrac{502}{3}\)
a; (5142 - 17 x 8 + 242 : 11) x (27 - 3 x 9)
= (5142 - 17 x 8 + 242 : 11) x (27 - 27)
= (5142 - 17 x 8 + 242 : 11) x 0
= 0
b;
(1 + \(\dfrac{1}{2}\)) \(\times\) (1 + \(\dfrac{1}{3}\)) \(\times\) ( 1 + \(\dfrac{1}{4}\)) \(\times\) ... \(\times\) (1 + \(\dfrac{1}{2010}\)) \(\times\)(1 + \(\dfrac{1}{2011}\))
= \(\dfrac{2+1}{2}\) \(\times\) \(\dfrac{3+1}{3}\) \(\times\) \(\dfrac{4+1}{4}\)\(\times\) ... \(\times\) \(\dfrac{2010+1}{2010}\)\(\times\) \(\dfrac{2011+1}{2011}\)
= \(\dfrac{3}{2}\)\(\times\)\(\dfrac{4}{3}\)\(\times\)\(\dfrac{5}{4}\)\(\times\)...\(\times\)\(\dfrac{2011}{2010}\)\(\times\)\(\dfrac{2012}{2011}\)
= \(\dfrac{2012}{2}\)
= 1006
\(\frac{31}{95}\)<\(\frac{1}{3}\)
\(\frac{1}{3}=\frac{2012}{6035}\)<
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:
\(A=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{15\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{303}{308}\\ \Rightarrow\dfrac{x-2}{x+3}=\dfrac{305-2}{305+3}\\ \Rightarrow x=305\)
333^444 = 111^444 . 3^444 = 111^444 . 81^111 > 8^111 . 111^444
`#3107.101107`
`a)`
Ta có:
\(\dfrac{2727}{3131}=\dfrac{2727\div27}{3131\div31}=\dfrac{27}{31}\)
Vì \(\dfrac{27}{31}=\dfrac{27}{31}\)
\(\Rightarrow\dfrac{27}{31}=\dfrac{2727}{3131}\)
`b)`
Ta có:
\(\dfrac{11}{31}=1-\dfrac{20}{31}=1-\dfrac{200}{310}\)
\(\dfrac{111}{311}=1-\dfrac{200}{311}\)
Vì \(\dfrac{200}{310}>\dfrac{200}{311}\)
\(\Rightarrow1-\dfrac{200}{310}< 1-\dfrac{200}{311}\)
\(\Rightarrow\dfrac{11}{31}< \dfrac{111}{311}.\)
27/31 = 2727/3131
11/31 bé hơn 111/311