\(\dfrac{2000}{2001}+\dfrac{1}{2001}=1\) mà \(\dfrac{1}{2001}< \dfrac{1}{2}\) nên \(\dfrac{2000}{2001}>\dfrac{1}{2}\)
\(\dfrac{2001}{2002}+_{ }\dfrac{1}{2002}=1\) mà \(\dfrac{1}{2002}< \dfrac{1}{2}\)nên \(\dfrac{2001}{2002}>\dfrac{1}{2}\)
- Hay A>1
- mà 2000+2001<2001+2002 nên B< 1
=> A>B