minh lam duoc roi . cach viet phan so ban bam vao o mau vang o cuoi trang .cu di con chuot xuong cuoi trang thi thay 1 o vang , vao xem huong dan la biet ngay ma.

2002/2001>:,2003/2002>1.....

CÓ 8 PHÂN SỐ MỖI PHÂN SỐ CÓ GIÁ TRỊ LỚN HƠN 1 VÂY TỔNG CỦA 8 PHÂN SỐ LỚN HƠN 1 SẼ LỚN HƠN 8.

a)2012/2013<2013/2014

b)2003x........< 2007x......

cho mình nha

\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>\frac{2001}{2001}+\frac{2002}{2002}+\frac{2003}{2003}+\frac{2004}{2004}+\frac{2005}{2005}+\frac{2006}{2006}+\frac{2007}{2007}+\frac{2008}{2008}\)

\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>1+1+1+1+1+1+1+1\)\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>8\)

\(A>8\)

a/

$A-3=\frac{2003}{2004}+\frac{2004}{2005}+\frac{2005}{2003}-3$

$=(1-\frac{1}{2004})+(1-\frac{1}{2005})+(1+\frac{2}{2003})-3$

$=\frac{2}{2003}-\frac{1}{2004}-\frac{1}{2005}$

$=(\frac{1}{2003}-\frac{1}{2004})+(\frac{1}{2003}-\frac{1}{2005})$

$>0+0=0$

$\Rightarrow A>3$

b/

$B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{2015^2}$

$< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}$

$=1-\frac{1}{2015}<1$

=1+1/2001+1+1/2002+1+1/2003+...+1+1/2008=8+1/2001+1/2002+1/2003+...+1/2008>8

\(\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>8\)

a)A = B

b)A>B

bạn ơi , phải giải thích chứ sao mà hiểu được

$A=\frac{{2005}^{2005}+1}{{2005}^{2006}+1}<\frac{{2005}^{2005}+1+2004}{{2005}^{2006}+1+2004}=\frac{2005.({2005}^{2004}+1)}{2005.({2005}^{2005}+1)}==\frac{{2005}^{2004}+1}{{2005}^{2005}+1}=B.$

Vậy A < B

$A=\frac{{2005}^{2005}+1}{{2005}^{2006}+1}<\frac{{2005}^{2005}+1+2004}{{2005}^{2006}+1+2004}=\frac{2005.({2005}^{2004}+1)}{2005.({2005}^{2005}+1)}==\frac{{2005}^{2004}+1}{{2005}^{2005}+1}=B.$

Vậy A < B