A=2001/2002+2002/2003

B=2001/2002+2003+2002/2002+2003

(tớ tách B ra đấy)

mà 2001//2002+2002/2003>2001/2002+2003+ 202/2002+2003

A>B

ta có \(\frac{2000+2002}{2001+2003}\)= \(\frac{2000}{2001+2003}\)+ \(\frac{2002}{2001+2003}\)=\(\frac{2000}{4004}\)+\(\frac{2002}{4004}\)

ta có \(\frac{2000}{2001}\)>\(\frac{2000}{4004}\) và \(\frac{2002}{2003}\)> \(\frac{2002}{4004}\)

 nên \(\frac{2000}{2001}\)+\(\frac{2002}{2003}\)>\(\frac{2000}{4004}\)+\(\frac{2002}{4004}\)

vậy \(\frac{2000}{2001}\)+\(\frac{2002}{2003}\)>\(\frac{2000+2002}{2001+2003}\)

\(\frac{2000+2002}{2001+2003}=\frac{2000}{2001+2003}+\frac{2002}{2001+2003}< \frac{2000}{2001}+\frac{2002}{2003}\)

mỗi  số hạng trong biểu thức A đều nhỏ hơn 1 mà có 15 số nên tổng A sẽ nhỏ hơn 15

ta thay tong tren <1+1+1+1+1+1+1+1+1+1+1+1+1+1+1

hay tong tren be hon 15

2001/2002=1-1/2002

2002/2003=1-1/2003

vi 1/2003<1/2002 nen 2001/2002<2002/2003

Ta có: 2003 x 2001 < 2002 x 2002

=> \(\frac{2001}{2002}\)<\(\frac{2002}{2003}\)

2 phân số trên: 2001/2002 = 2002/2003

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

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

Ta có:

2002/2001=1+1/2001

2003/2002=1+1/2002

2004/2003= 1+ 1/2003

2005/2004= 1+ 1/2004

2006/2005=1+ 1/2005

2007/2006= 1+ 1/2006

2008/2007=1 + 1/2007.

2009/2008=1+ 1/2008.

=> 2002/2001+2003/2002+2004?2003+2005/2004+2006/2005+ 2007/2006+ 2008/2007+ 2009/2008= 1+1+1+1+1+1+1+1+1/2001+1/2002+1/2003+1/2004+1/2005+1/2006+1/2007+1/2008>8.

Nhớ k đúng cho mình nha!! Thanks!!!

Tham khảo : 

Sứa , san hô , hải quỳ , thủy tức , sứa tu dài ,...

\(\dfrac{2001+2002}{2002+2003}< \dfrac{2001}{2002}+\dfrac{2002}{2003}\)

ta thấy:

\(B< 1\Rightarrow B< \frac{10^{2002}+1+9}{10^{2003}+1+9}=\frac{10^{2002}+10}{10^{2003}+10}=\frac{10\left(10^{2001}+1\right)}{10\left(10^{2002}+1\right)}=\frac{10^{2001}+1}{10^{2002}+1}=A\)

=>B<A

vậy.......

Ta có:

\(A=\frac{10^{2001}+1}{10^{2002}+1}\Rightarrow10A=\frac{10\left(10^{2001}+1\right)}{10^{2002}+1}=\frac{10^{2002}+10}{10^{2002}+1}=\frac{10^{2002}+1+9}{10^{2002}+1}=1+\frac{9}{10^{2002}+1}\)

\(B=\frac{10^{2002}+1}{10^{2003}+1}\Rightarrow10B=\frac{10\left(10^{2002}+1\right)}{10^{2003}+1}=\frac{10^{2003}+10}{10^{2003}+1}=\frac{10^{2003}+1+9}{10^{2003}+1}=1+\frac{9}{10^{2003}+1}\)

Vì \(\frac{9}{10^{2002}+1}>\frac{9}{2^{2003}+1}\Rightarrow1+\frac{9}{10^{2002}+1}>1+\frac{9}{2^{2003}+1}\Rightarrow10A>10B\Rightarrow A>B\)

Vậy A > B

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