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ta có :
\(\frac{2008}{2009}< 1\)
\(\frac{2009}{2010}< 1\)
\(\frac{2010}{2011}< 1\)
\(\frac{2011}{2012}< 1\)
\(\Rightarrow\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}< 1+1+1+1\)
\(\Rightarrow\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2012}< 4\)
vậy ....................
2008/2009+2009/2010+2010/2011+2011/2008 4
=2008/2008=1 4
Vì 1<4 nên 2008/2009+2009/2010+2011/2008 < 4
A = [2 + (-3)] + [4 + (-5)] +.....+ [2008 + (-2009)] + [2010 + (-2011)]
= (-1) + (-1) +.....+ (-1) + (-1) (Có 2010 :2 = 1005 số -1)
= 1005. (-1) = -1005
2 + ( -3 ) + 4 (-5 ) + ... + 2008 + ( -2009 ) + 2010 + (-2011 ) + 2012
= [2 + ( -3 )] + [4 (-5 )] + ... + [2008 + ( -2009 )] + [2010 + (-2011)] + 2012 (có 1005 cặp)
= (-1) + (-1) + ... + (-1) + 2012
= (-1) . 1005 +2012
= (-1005) + 2012
= 1007
2 + (-3) + 4 + (-5) +..........+ 2008 + (-2009) + 2010 + (-2011) + 2012=-2010
2 + (-3) + 4 + (-5) +..........+ 2008 + (-2009) + 2010 + (-2011) + 2012
= 1007
Ta có dãy số : 1,2,3,...,2011,2012 có : (2012-1)+1=2012(số)
Ta thấy : 2012 :2 nên :
[(-1)+2]+[(-3)+4]+...+[(-2009)+2010]+[(-2011)+2012] (có 1006 cặp)
= 1 . 1006
=1006
Đặt cả tổng đó là A(hayf j đó tùy bạn)
ta có
A=1+2+3+4+...+2012-2(1+3+5+...+2011)
A=2025078-2.1012036
A=2025078-2024072
A=1006
*xong*
Ta có 2008/2009 < 1; 2009/2010 < 1; 2010/2011 < 1; 2011/2012 < 1
Nên : 2008/2009 + 2009/2010 + 2010/2011 + 2011/2012 < 1 + 1 + 1 + 1
Ta có 2008/2009 < 1; 2009/2010 < 1; 2010/2011 < 1; 2011/2012 < 1
Nên : 2008/2009 + 2009/2010 + 2010/2011 + 2011/2012 < 1 + 1 + 1 + 1
Hay A < 4
A < B
\(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2008}=1-\frac{1}{2009}+1-\frac{1}{2010}+1-\frac{1}{2011}+1+\frac{3}{2008}=1+1+1+1+\frac{1}{2008}+\frac{1}{2008}+\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}=4+\left(\frac{1}{2008}-\frac{1}{2009}\right)+\left(\frac{1}{2008}-\frac{1}{2010}\right)+\left(\frac{1}{2008}-\frac{1}{2011}\right)\left(vì:2008>2009>2010>2011\right)\Rightarrow\frac{1}{2008}>\frac{1}{2009}>\frac{1}{2010}>\frac{1}{2011}\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{2008}-\frac{1}{2009}>0\\\frac{1}{2008}-\frac{1}{2010}>0\\\frac{1}{2008}-\frac{1}{2011}>0\end{matrix}\right.\Rightarrow4+\left(\frac{1}{2008}-\frac{1}{2009}\right)+\left(\frac{1}{2008}-\frac{1}{2010}\right)+\left(\frac{1}{2008}-\frac{1}{2011}\right)>4+0+0+0=4\Rightarrow\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2008}>4\)
2+(-3)+4+...+2012
=2-3+4-5+....+2008-2009+2010-2011+2012
=-1+(-1)+(-1)+(-1)+(-1)+...+(-1)+(-1)+2012
=-1005+2012
=1007
Nhớ k nhe