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\(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2006}\)
\(A=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{2}{2006}\)
\(A=\left(1+1+1\right)+\left(\frac{1}{2006}-\frac{1}{2007}\right)+\left(\frac{1}{2006}-\frac{1}{2008}\right)\)
\(A=3+\left(\frac{1}{2006}-\frac{1}{2007}\right)+\left(\frac{1}{2006}-\frac{1}{2008}\right)\)
Ta thấy : \(\frac{1}{2006}-\frac{1}{2007}>0\); \(\frac{1}{2006}-\frac{1}{2008}>0\)\(\Rightarrow A>3\)
\(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2006}=1-\frac{1}{2007}+1-\frac{1}{2008}+1+\frac{2}{2006}=3+\left(\frac{1}{2006}-\frac{1}{2007}\right)+\left(\frac{1}{2006}-\frac{1}{2008}\right)\)
Vì 2006<2007, 2006<2008 nên \(\frac{1}{2006}>\frac{1}{2007};\frac{1}{2006}>\frac{1}{2008}=>\frac{1}{2006}-\frac{1}{2007}>0,\frac{1}{2006}-\frac{1}{2008}>0\)
=> \(A=3+\left(\frac{1}{2006}-\frac{1}{2007}\right)+\left(\frac{1}{2006}-\frac{1}{2008}\right)>3=>A>3\)
bạn ơi cái dấu bằng to và dấu lớn to là dấu suy ra ak
\(\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2006}\)
\(\Rightarrow\frac{2008}{2006}>1\)
\(\frac{2006}{2007}< 1;\frac{2007}{2008}< 1\)
\(\Rightarrow\frac{2006}{2007}+\frac{2007}{2008}< 2\)
\(\Rightarrow\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2006}< 3\)
A =2006/2007+2007/2008+2008/2006
= \(\frac{2006}{2007}\)+ \(\frac{2007+1}{2008}\)+ \(\frac{2008}{2006+2}\)
= 1 - \(\frac{1}{2007}\)+ 1 - \(\frac{1}{2008}\)+ 1 + \(\frac{1}{2006}\)+ \(\frac{1}{2006}\)
= 3 + ( \(\frac{1}{2006}\)- \(\frac{1}{2007}\)) + ( \(\frac{1}{2006}\)- \(\frac{1}{2008}\))
vì \(\frac{1}{2006}\)> \(\frac{1}{2007}\), \(\frac{1}{2006}\)> \(\frac{1}{2008}\)nên A > 3
Ta có: 3 = 1 + 1 + 1
Ta có: 2006/2007 < 1 ; 2007/2008 < 1 ; 2008/2009 < 1
Nên 2006/2007 + 2007/2008+ 2008/2009 < (1+1+1=3)
Ta có: \(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2006}\)
\(A=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{2}{2006}\)
\(A=3+\frac{1}{2006}-\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2008}>3\)
Vậy A > 3
2006/2007<1
2007/2008<1
2008<2009<1
2009/2006>1
A=2006/2007+2007/2008+2008/2009+2009/2006\(\approx\)3+1=4
ta thấy: 2008/2006>1
=>2006/2007+2007/2008+2008/2006>1
Vậy 2006/2007+2007/2008+2008/2006>1
***nhé
\(A=\frac{2006+2007}{2006.2007}=\frac{2006}{2006.2007}+\frac{2007}{2006.2007}=\frac{1}{2007}+\frac{1}{2006}\)
\(B=\frac{2007+2008}{2007.2008}=\frac{2007}{2007.2008}+\frac{2008}{2007.2008}=\frac{1}{2008}+\frac{1}{2007}\)
Vì \(\frac{1}{2007}+\frac{1}{2006}>\frac{1}{2008}+\frac{1}{2007}\)
=> \(A>B\)
a) Ta có:
\(1-\frac{2005}{2006}=\frac{1}{2006}\)
\(1-\frac{2006}{2007}=\frac{1}{2007}\)
Vì \(\frac{1}{2006}>\frac{1}{2007}\)nên \(\frac{2005}{2006}>\frac{2006}{2007}\)
b) Ta có:
\(\frac{2008}{2007}-1=\frac{1}{2007}\)
\(\frac{2007}{2006}-1=\frac{1}{2006}\)
Vì \(\frac{1}{2006}>\frac{1}{2007}\)nên \(\frac{2008}{2007}< \frac{2007}{2006}\)
a, \(\frac{2005}{2006}v\text{à}\frac{2006}{2007}\)= \(\frac{2005\cdot2007}{2006\cdot2007}\)và \(\frac{2006\cdot2006}{2007\cdot2006}\)
= \(\frac{4024035}{4026042}\)< \(\frac{4024036}{4026042}\)
b, \(\frac{2008}{2007}v\text{à}\frac{2007}{2006}\)= \(\frac{2008\cdot2006}{2007\cdot2006}v\text{à}\frac{2007\cdot2007}{2006\cdot2007}\)
=\(\frac{4028048}{4026042}\)< \(\frac{4028049}{4026042}\)
2.006/2.007 + 2.007/2.008 < 2006 + 2.007/2.007 + 2.008
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