Đề là tính A, B hả em.
[TEX]A= \frac{ \frac{1}{2}+ \frac{1}{3}+...+ \frac{1}{2000}}{ \frac{1999}{1}+ \frac{1998}{2}+ ...+ \frac{1}{1999}}[/TEX]
Xét mẫu: [TEX]\frac{1999}{1}+ \frac{1998}{2}+ ...+ \frac{1}{1999}= 1999+ \frac{2000-2}{2}+ \frac{2000-3}{3}+...+ \frac{2000-1999}{1999}=1999+ 2000 \times \left( \frac{1}{2}+ \frac{1}{3}+...+ \frac{1}{1999} \right) - 1998 = 1+ 2000 \times \left( \frac{1}{2}+ \frac{1}{3}+...+ \frac{1}{1999} \right) = 2000 \times \left( \frac{1}{2}+ \frac{1}{3}+...+ \frac{1}{2000} \right)[/TEX].
Vậy [TEX]\fbox{A= \frac{1}{2000}}[/TEX].
[TEX]B= \left( 1+ \frac{1}{1.3} \right) \left(1+ \frac{1}{2.4} \right) \left( 1+ \frac{1}{3.5} \right) ... \left(1+ \frac{1}{99.101} \right)[/TEX]
[TEX]= \frac{4}{1.3}. \frac{9}{2.4}. \frac{16}{3.5}... \frac{10000}{99.101}[/TEX]
[TEX]= \frac{2.2.3.3.4.4...100.100}{(1.3).(2.4).(3.5)...(99.101)}[/TEX]
[TEX]= \frac{(2.3.4...100).(2.3.4...100)}{(1.2...99).(3.4...101)}[/TEX]
[TEX]= \frac{100.2}{101} = \fbox{ \frac{200}{101}}[/TEX].