NP
1
K
Khách
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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:
\(\dfrac{2014.2015-1}{2013.2015+2014}\)
\(=\dfrac{\left(2013+1\right).2015-1}{2013.2015+2014}\)
\(=\dfrac{2013.2015+2015-1}{2013.2015+2014}\)
\(=\dfrac{2013.2015+2014}{2013.2015+2014}\)
\(=1\)
VB
27 tháng 1 2019
1.
ta có: 2009A= (2009^2010+ 2009)/ (2009^2010+1)= (2009^10+1+2008)/(2009^2010+1)=1+ [2008/(2009^2010+1)]
làm tương tự như trên ta được :
2009B=1-[4016/(2009^2011-2)]
lại có:
2009A= .............(nt) > 1
2009B=...........<1
=>2009A>2009B
=>A>B
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 ^^
30 tháng 6 2017
\(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{x\left(x+2\right)}=1\dfrac{2009}{2011}\)
\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{4020}{4022}\)
\(\Rightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{4020}{4022}\)
\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{4020}{4022}\)
\(\Rightarrow1-\dfrac{1}{x+2}=\dfrac{4020}{4022}\)
\(\Rightarrow\dfrac{1}{x+2}=\dfrac{1}{2011}\)
\(\Rightarrow x+2=2011\Rightarrow x=2009\)
Vậy x = 2009
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 ~
LG
12 tháng 7 2017
Đặt \(A=\dfrac{2009^{2008}+1}{2009^{2009}+1}\) và \(B=\dfrac{2009^{2007}+1}{2009^{2008}+1}\)
Ta có:
\(2009A=\dfrac{2009.\left(2009^{2008}+1\right)}{2009^{2009}+1}=\dfrac{2009^{2009}+2009}{2009^{2009}+1}\)
\(=\dfrac{2009^{2009}+1+2008}{2009^{2009}+1}=\dfrac{2009^{2009}+1}{2009^{2009}+1}+\dfrac{2008}{2009^{2009}+1}\)
\(=1+\dfrac{1}{2009^{2009}+1}\)
\(2009B=\dfrac{2009.\left(2009^{2007}+1\right)}{2009^{2008}+1}=\dfrac{2009^{2008}+2009}{2009^{2008}+1}\)
\(=\dfrac{2008^{2008}+1+2008}{2009^{2008}+1}=\dfrac{2008^{2008}+1}{2009^{2008}+1}+\dfrac{2008}{2009^{2008}+1}\)
\(=1+\dfrac{2008}{2009^{2008}+1}\)
Vì \(1+\dfrac{2008}{2009^{2009}+1}< 1+\dfrac{2008}{2009^{2008}+1}\)
Nên \(10A< 10B\) \(\Rightarrow A< B\)
Vậy \(\dfrac{2009^{2008}+1}{2009^{2009}+1}< \dfrac{2009^{2007}+1}{2009^{2008}+1}\)
~ Học tốt ~
12 tháng 7 2017
Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(A=\dfrac{2009^{2008}+1}{2009^{2009}+1}< 1\)
\(\Rightarrow A< \dfrac{2009^{2008}+1+2008}{2009^{2009}+1+2008}\Rightarrow A< \dfrac{2009^{2008}+2009}{2009^{2009}+2009}\Rightarrow A< \dfrac{2009\left(2009^{2007}+1\right)}{2009\left(2009^{2008}+1\right)}\Rightarrow A< \dfrac{2009^{2007}+1}{2009^{2008}+1}=B\)\(\Rightarrow A< B\)
17 tháng 3 2018
2, ta thấy:
\(\dfrac{2008}{2009}< \dfrac{2008}{2009+2010}\left(1\right)\)
\(\dfrac{2009}{2010}< \dfrac{2009}{2009+20010}\left(2\right)\)
từ (1) và (2) cộng vế với vế ta đc :\(\dfrac{2008}{2009}+\dfrac{2009}{20010}< \dfrac{2008}{2009+2010}+\dfrac{2009}{2009+2010}=\dfrac{2008+2009}{2009+2010}\)
NU
4
P
27 tháng 4 2018
Ta có:
\(\dfrac{2010}{2009} > 1\)
\(\dfrac{2009}{2010} < 1\)
=> \(\dfrac{2010}{2009} >\dfrac{2009}{2010}\)
27 tháng 4 2018
2010/2009=2010*2010/2009*2010
2009/2010=2009*2009/2009*2010
mà 2010*2010/2009*2010>2009*2009/2009*2010
=> 2010/2009>2009/2010
mk ko biết viết dấu phân số
5 tháng 8 2018
a, (x + 1) + (x + 4) + ... + (x + 28) = 155
x + 1 + x + 4 + ... + x + 28 = 155
(x + x + x + ... + x) + (1 + 4 + ... + 28) = 155
x . 10 + 145 = 155
x . 10 = 155 - 145
x . 10 = 10
x = 10 : 10
x = 1
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 ! ^_^ !