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

 \(A\approx4\)

\(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{1}{2009}\)

\(=3-\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}>1\).

\(B=\frac{2006+2007+2008}{2007+2008+2009}< \frac{2007+2008+2009}{2007+2008+2009}=1\).

Suy ra \(A>B\).

Ta có : \(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2006}\)

                \(=\frac{2007-1}{2007}+\frac{2008-1}{2008}+\frac{2009-1}{2009}+\frac{2006+3}{2006}\)

                  \(=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{1}{2009}+1+\frac{3}{2006}\)

                  \(=\left(1+1+1+1\right)-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}\right)\)

                  \(< 4-\left(\frac{1}{2009}+\frac{1}{2009}+\frac{1}{2009}-\frac{3}{2009}\right)\)     

                    \(=4\)

=> A < 4 

Vậy A < 4 

câu này của bà Hồ Thu Giang nè kêu bả chỉ

Ta có A= (1 -1/2007) +(1-1/2008)+(1-1/2009)+(1+3/2006)= 4-(1/2007+1/2008+1/2009-3/2006) <4

ta có: \(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2006}\)

A = \(1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{1}{2009}+1+\frac{3}{2006}\)

A= \(4\)\(+\frac{3}{2006}-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)\)

Do 1/2007 < 1/2006 ; 1/2008<1/2006 ; 1/2009<1/2006=> 1/2007 + 1/2008 + 1/2009 < 1/2006 + 1/2006 + 1/2006

Mà 1/2006 + 1/2006 + 1/2006 = 3/2006

=> 3/2006  -( 1/2007 + 1/2008 + 1/2009) > 0 

=> \(4+\frac{3}{2006}-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)>4\)

=> A > 4

Ta có:\(\frac{2006}{2007}< 1\)

           \(\frac{2007}{2008}< 1\)

           \(\frac{2008}{2009}< 1\)

            \(\frac{2009}{2006}>1\)\(\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2006}< 4\)

Mk chưa thấy ai làm bài sai như thế đấy lỗi đó thì hs lớp 4 cũng phát hiện ra

bạn vào câu hỏi tương tự ấy

\(\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2006}\)

\(=\left(1-\frac{1}{2007}\right)+\left(1-\frac{1}{2008}\right)+\left(1-\frac{1}{2009}\right)+\left(1+\frac{3}{2006}\right)\)

\(=\left(1+1+1+1\right)-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)+\frac{3}{2006}\)

\(< 4-\left(\frac{1}{2006}+\frac{1}{2006}+\frac{1}{2006}\right)+\frac{3}{2006}\)

\(=4-\frac{3}{2006}+\frac{3}{2006}\)

\(=4\)

\(\Rightarrow\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2006}< 4\)

Gọi biểu thức trên là A

\(\frac{2006}{2007}=1-\frac{1}{2007}\)

\(\frac{2007}{2008}=1-\frac{1}{2008}\)

\(\frac{2008}{2009}=1-\frac{1}{2009}\)

\(\frac{2009}{2006}=1+\frac{3}{2006}\)

\(\Rightarrow A=4-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}\right)\)

\(\Rightarrow A< 4\)

Study well