NV
0
K
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TD
0
15 tháng 5 2017
\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>\frac{2001}{2001}+\frac{2002}{2002}+\frac{2003}{2003}+\frac{2004}{2004}+\frac{2005}{2005}+\frac{2006}{2006}+\frac{2007}{2007}+\frac{2008}{2008}\)
\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>1+1+1+1+1+1+1+1\)\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>8\)
\(A>8\)
27 tháng 2 2022
a, Ta có: \(\dfrac{27}{37}< \dfrac{27}{18};\dfrac{27}{18}< \dfrac{28}{18}\Rightarrow\dfrac{27}{37}< \dfrac{28}{18}\)
b, Ta có: \(1-\dfrac{2003}{2005}=\dfrac{2}{2005}\)
\(1-\dfrac{2001}{2003}=\dfrac{2}{2003}\)
Vì \(\dfrac{2}{2005}< \dfrac{2}{2003}\Rightarrow\dfrac{2003}{2005}>\dfrac{2001}{2003}\)
TQ
8
15 tháng 8 2017
a, 2013/2018 < 2012/2018
b, 2013/2008 < 2008/2003
c,24/47 > 13/27
d,37/23 < 42/22
e 1/2 > 1/2017
g, 12/13 > 6/7
DQ
2
12 tháng 7 2019
2003/2004 + 2004/2005 + 2005/2003
= 1 - 1/2004 + 1 - 1/2005 + 1 + 1/2003 + 1/2003
=(1+1+1)-(1/2004 - 1/2003 + 1/2005 - 1/2003)
= 3 - (1/2004 - 1/2003 + 1/2005 - 1/2003)
Vì 1/2004 < 1/2003 ; 1/2005 < 1/2003
=>1/2004 - 1/2003 + 1/2005 - 1/2003 < 0
=> 3 - (...) > 3
Vậy. ...
K mình nha
29 tháng 7 2015
a) \(\frac{2005.2007-1}{2004+2005.2006}=\frac{\left(2014+1\right).2007-1}{2004+2005.2006}=\frac{2004+2005.2007-1}{2004+2005-2006}=\frac{2004+2005.2006}{2004+2005.2006}=1\)
DQ
2
DD
12 tháng 7 2019
\(\frac{2003}{2004}+\frac{2004}{2005}+\frac{2005}{2003}=1-\frac{1}{2004}+1-\frac{1}{2005}+1+\frac{2}{2003}\)
\(=3+\left(\frac{1}{2003}-\frac{1}{2004}\right)+\left(\frac{1}{2003}-\frac{1}{2005}\right)\)
Do \(\frac{1}{2003}>\frac{1}{2004}>\frac{1}{2005}.\) nên \(\left(\frac{1}{2003}-\frac{1}{2004}\right)+\left(\frac{1}{2003}-\frac{1}{2005}\right)>0\)
Vì vậy \(3+\left(\frac{1}{2003}-\frac{1}{2004}\right)+\left(\frac{1}{2003}-\frac{1}{2005}\right)>3\) (đpcm)
12 tháng 7 2019
\(A=\frac{2003}{2004}+\frac{2004}{2005}+\frac{2005}{2003}\)
\(=(1-\frac{1}{2004})+(1-\frac{1}{2005})+(1+\frac{2}{2003})\)
\(=3+(\frac{1}{2003}+\frac{1}{2003}-\frac{1}{2004}-\frac{1}{2005})\)
Do\(\frac{1}{2003}\)>\(\frac{1}{2004}\)>\(\frac{1}{2005}\)
\(\Rightarrow\frac{1}{2003}+\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\)>\(0\)
\(\Rightarrow3+(\frac{1}{2003}-\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2005})\)>\(3\)
\(\Rightarrow A\)>\(3\)
TK
a) 2005*2007-1 b)2003*2004+2005*10+1994
2004+2005*2006 2005*2004-2003*2004
c) ( 5+3/8+18+1/2-7-5/24 )
3
10 tháng 4 2016
c) 5 + \(\frac{3}{8}\)+18 + \(\frac{1}{2}\) - 7 - \(\frac{5}{24}\)
=\(\frac{43}{8}\)+ \(\frac{35}{2}\) +\(\frac{163}{24}\)
=\(\frac{129}{24}\)+ \(\frac{420}{24}\)+\(\frac{163}{24}\)
= \(\frac{58}{51}\)
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