\(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2006}\)

\(A=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{2}{2006}\)

\(A=\left(1+1+1\right)+\left(\frac{1}{2006}-\frac{1}{2007}\right)+\left(\frac{1}{2006}-\frac{1}{2008}\right)\)

\(A=3+\left(\frac{1}{2006}-\frac{1}{2007}\right)+\left(\frac{1}{2006}-\frac{1}{2008}\right)\)

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

Mình nhé bạn :  

http://olm.vn/thanhvien/ngathongminh

2.006/2.007 + 2.007/2.008 < 2006 + 2.007/2.007 + 2.008

Chúc bạn học tốt.

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hong be oi

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

 \(A\approx4\)

Tham khảo:

em cảm ơn ạ

ta có A= 3 - (1/2006 + 1/2007) + 2/2006 
để so sánh A với 3 ta sẽ chuyển về so sánh (1/2006 + 1/2007) với 2/2006 
nếu (1/2006 + 1/2007) < hơn 2/2006 thì A > 3 và ngược lại 
So sánh: ta có (1/2006+1/2007) < (1/2006+1/2006)=2/2006 
vậy A>3

A= (1-1/2007)+(1-2008)+(1+2/2006) 
=3+1/2006+1/2006-1/2007-1/2008 
Mà 1/2006>1/2007>1/2008 nên A>3