K
Khách
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Các câu hỏi dưới đây có thể giống với câu hỏi trên
29 tháng 3 2018
Ta có: 2004.2007=2004.(2004+3)=2004.2004+3.2004
2005.2006=(2004+1).(2004+2)=2004.2004+3.2004+2
Vì 2004.2004+3.2004 < 2004.2004+3.2004 +2 nên 2004.2007 < 2005.2006
29 tháng 3 2018
Ta có 2004.2007 = ( 2005 - 1 ) . 2007
= 2005 . 2007 - 2007 (1)
Ta lại có 2005 . 2006 = 2005 . ( 2007 - 1 )
= 2005 . 2007 - 2005 (2)
Vì 2005 < 2007 nên từ (1) và (2) => 2005 . 2007 - 2005 > 2005 . 2007 - 2007
=> 2005 . 2006 > 2004 . 2007
KT
29 tháng 3 2018
\(2004.2007=2004.\left(2006+1\right)\)
\(=2004.2006+2004\)
\(2005.2006=\left(2004+1\right).2006\)
\(=2004.2006+2006\)
do \(2004< 2006\)
nên \(2004.2006+2004< 2004.2006+2006\)
hay \(2004.2007< 2005.2006\)
HN
29 tháng 3 2018
2004 . 2007 = 2004 . ( 2006 + 1 ) = 2004 . 2006 + 2004
2005 . 2006 = ( 2004 + 1 ) . 2006 = 2004 . 2006 + 2006
Vì 20004 . 2006 + 2004 < 2004 . 2006 + 2006 nên 2004 . 2007 < 2005 . 2006
Vậy 2004 . 2007 < 2005 . 2006
4 tháng 7 2017
\(A=\dfrac{1995.1994-1}{1993.1995+1994}=\dfrac{1995\left(1993+1\right)-1}{1993.1995+1994}=\dfrac{1995.1993+1995-1}{1993.1995+1994}=\dfrac{1995.1993+1994}{1995.1993-1994}=1\)\(B=\dfrac{2004.2004+3006}{2005.2005-1003}=\dfrac{2004.2004+2004.1+1002}{2005.2005-1003}=\dfrac{2004.2005+1002}{2005.2005-1003}=\dfrac{2004.2005+1002}{2004.2005+2005-1003}=\dfrac{2004.2005+1002}{2004.2005+1002}=1\)\(C=\dfrac{2010.2011-1}{2009.2011+2010}=\dfrac{2009.2011+2011-1}{2009.2011+2010}=\dfrac{2019.2011+2010}{2009.20011+2010}=1\)\(D=\dfrac{2014.2015-1}{2013.2015+2013}=\dfrac{2013.2015+2014-1}{2013.2015+2013}=\dfrac{2013.2015+2013}{2013.2015+2013}=1\)
LG
4 tháng 7 2017
Câu 1 nhầm đề nha bạn mình sửa:
\(\dfrac{1995.1994-1}{1993.1995+1994}\)
\(=\dfrac{1995.\left(1993+1\right)-1}{1993.1995+1994}\)
\(=\dfrac{1995.1993+1995-1}{1993.1995+1994}\)
\(=\dfrac{1993.1995+1994}{1993.1995+1994}\)
\(=1\)
Câu 2: \(\dfrac{2004.2004+3006}{2005.2005-1003}\)
\(=\dfrac{2004.2004+2004+1002}{\left(2004+1\right).\left(2004+1\right)-1003}\)
\(=\dfrac{2004.2004+2004+1002}{2004.2004+2004+1-1003}\)
\(=\dfrac{2004.2004+2004+1002}{2004.2004+2004+1002}\)
\(=1\)
Câu 3:\(\dfrac{2010.2011-1}{2009.2011+2010}\)
\(=\dfrac{\left(2009+1\right).2011-1}{2009.2011+2010}\)
\(=\dfrac{2009.2011+2011-1}{2009.2011+2010}\)
\(=\dfrac{2009.2011+2010}{2009.2011+2010}\)
= 1
Câu 4:Nhầm để, sửa:
\(\dfrac{2014.2015-1}{2013.2015+2014}\)
\(=\dfrac{\left(2013+1\right).2015-1}{2013.2015+2014}\)
\(=\dfrac{2013.2015+2015-1}{2013.2015+2014}\)
\(=\dfrac{2013.2015+2014}{2013.2015+2014}\)
\(=1\)
13 tháng 2 2019
=\(\dfrac{1}{2009.\left(\dfrac{1}{2009}+\dfrac{1}{2011}+\dfrac{1}{2010}\right)}+\dfrac{1}{2010.\left(\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2011}\right)}+\dfrac{1}{2011.\left(\dfrac{1}{2011}+\dfrac{1}{2009}+\dfrac{1}{2010}\right)}\)\(=\dfrac{1}{2009}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)+\dfrac{1}{2010}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)+\dfrac{1}{2011}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)\)
\(=\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right):\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)=1\)
4 tháng 5 2021
Giải:
Ta có:
A=20092008+1/20092009+1
2009A=20092009+2009/20092009+1
2009A=20092009+1+2008/20092009+1
2009A=20092009+1/20092009+1 + 2008/20092009+1
2009A=1+2008/20092009+1
Tương tự:
B=20092009+1/20092010+1
2009B=1+2008/20092010+1
Vì 2008/20092009+1 > 2008/20092010+1 nên 2009A>2009B
⇒A>B
26 tháng 2 2018
Ta có :
\(B=\frac{2008+2009+2010}{2009+2010+2011}=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
Vì :
\(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\)
\(\frac{2009}{2010}>\frac{2009}{2009+2010+2011}\)
\(\frac{2010}{2011}>\frac{2010}{2009+2010+2011}\)
Nên \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
\(\Rightarrow\)\(A>B\)
Vậy \(A>B\)
26 tháng 2 2018
Ta có: \(B=\frac{2008+2009+2010}{2009+2010+2011}\)
\(=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
Vì \(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\)
\(\frac{2009}{2010}>\frac{2009}{2009+2010+2011}\)
\(\frac{2010}{2011}>\frac{2010}{2009+2010+2011}\)
nên \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008+2009+2010}{2009+2010+2011}\)
hay A > B
Vậy A > B
S
26 tháng 2 2018
\(B=\dfrac{2008+2009+2010}{2009+2010+2011}=\dfrac{2008}{2009+2010+2011}+\dfrac{2009}{2009+2010+2011}+\dfrac{2010}{2009+2010+2011}\)Ta có : \(\dfrac{2008}{2009}>\dfrac{2008}{2009+2010+2011}\)
\(\dfrac{2009}{2010}>\dfrac{2009}{2009+2010+2011}\)
\(\dfrac{2010}{2011}>\dfrac{2010}{2009+2010+2011}\)\(=>\dfrac{2008}{2009}+\dfrac{2009}{2010}+\dfrac{2010}{2011}>\dfrac{2008+2009+2010}{2009+2010+2011}\)
Hay A > B
SM
2
MV
22 tháng 4 2017
B = \(\dfrac{2009^{2009}+1}{2009^{2010}+1}\)<\(\dfrac{2009^{2009}+1+2008}{2009^{2010}+1+2008}\)=\(\dfrac{2009^{2009}+2009}{2009^{2010}+2009}\)=\(\dfrac{2009.\left(2009^{2008}+1\right)}{2009.\left(2009^{2009}+1\right)}\)=\(\dfrac{2009^{2008}+1}{2009^{2019}+1}\)= A
Vậy A > B
22 tháng 4 2017
Ta có :
\(2009A=\dfrac{2009^{2009}+2009}{2009^{2009}+1}=\dfrac{2009^{2009}+1+2008}{2009^{2009}+1}=\dfrac{2009^{2009}+1}{2009^{2009}+1}+\dfrac{2008}{2009^{2009}+1}=1+\dfrac{2008}{2009^{2009}+1}\)
\(2009B=\dfrac{2009^{2010}+2009}{2009^{2010}+1}=\dfrac{2009^{2010}+1+2008}{2010^{2010}+1}=\dfrac{2009^{2010}+1}{2009^{2010}+1}+\dfrac{2008}{2009^{2010}+1}=1+\dfrac{2008}{2009^{2010}}\)
\(\)Vì \(1+\dfrac{2008}{2009^{2009}+1}>1+\dfrac{2008}{2009^{2010}+1}\Rightarrow A>B\)
~ Học tốt ~
15 tháng 3 2018
ta thấy:
\(\dfrac{2008}{2009}>\dfrac{2008}{2009+2010}\)(1)
\(\dfrac{2009}{2010}>\dfrac{2009}{2009+2010}\)(2)
từ 1 và 2 cộng vế với vế ta dc \(\dfrac{2008}{2009}+\dfrac{2009}{2010}>\dfrac{2008}{2009+2010}+\dfrac{2009}{2009+2010}=\dfrac{2008+2009}{2009+2010}\)
chúc bạn học tốt ^^
Xét tử số :
2004 . 2007 + 6
= (2005 - 1) . 2007 + 6
= 2005 . 2007 - 2007 + 6
= 2005 . (2005 + 2) - 2007 + 6
= 2005 . 2005 + 2005 . 2 - 2007 + 6
= 2005 . 2005 + 4010 - 2007 + 6
= 2005 . 2005 + 2009
Vậy ta có : \(\dfrac{2004.2007+6}{2005.2005+2009}=\dfrac{2005.2005+2009}{2005.2005+2009}=1\)
Thanks ban nhiu nha ! GOOD LUCK ! ^_^ !