Dễ thấy:

\(\frac{2008}{2009}>\frac{2008}{2009+2010+2011}\)

\(\frac{2009}{2010}>\frac{2009}{2009+2010+2011}\)

\(\frac{2010}{2011}>\frac{2010}{2009+2010+2011}\)

=>\(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)

Hay \(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}>\frac{2008+2009+2010}{2009+2010+2011}\)

Vậy A > B

\(\frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}+\frac{2011}{2008}=1-\frac{1}{2009}+1-\frac{1}{2010}+1-\frac{1}{2011}+1+\frac{3}{2008}\)

\(=4+\left(\frac{1}{2008}-\frac{1}{2009}\right)+\left(\frac{1}{2008}-\frac{1}{2010}\right)+\left(\frac{1}{2008}-\frac{1}{2011}\right)>4\)

4 bé hơn

sẽ bằng nhau 

câu hỏi tương tự

Ta có :

\(N=\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}\)

\(=\frac{2009^{2010}+2009}{2009^{2011}+2009}=\frac{2009.\left(2009^{2009}+1\right)}{2009.\left(2009^{2010}+1\right)}\)

\(=\frac{2009^{2009}+1}{2009^{2010}+1}=M\)

Vậy \(M>N\)

Ta có: \(B< 1\)

\(\Rightarrow B< \frac{2009^{2010}-2+3}{2009^{2011}-2+3}=\frac{2009^{2010}+1}{2009^{2011}+1}\left(1\right)\)

Mà \(\frac{2009^{2010}+1}{2009^{2011}+1}< 1\)

\(\Rightarrow\frac{2009^{2010}+1}{2009^{2011}+1}< \frac{2009^{2010}+1+2008}{2009^{2011}+1+2008}=\frac{2009^{2010}+2009}{2009^{2011}+2009}=\frac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\frac{2009^{2009}+1}{2009^{2010}+1}=A\left(2\right)\)

Từ (1) và (2) suy ra A > B

Ta có :

$B=\genfrac{}{}{0ex}{}{2009^{2010}-2}{2009^{2011}-2}<1$

$\iff B<\genfrac{}{}{0ex}{}{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\genfrac{}{}{0ex}{}{2009^{2010}+2009}{2009^{2011}+2009}=\genfrac{}{}{0ex}{}{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\genfrac{}{}{0ex}{}{2009^{2009}+1}{2009^{2010}+1}=A$

$\iff A>B$

Bạn Edogawa giải thích rõ hơn cho mình hiểu được không?

dễ quá cái này so sánh B với 1 sau đó suy ra B< B- thêm tử và mẫu 2011