(1 1999/1)(1 1999/2)...(1 1999/1000)/(1 1000/1)(1 1000/2)...(1 1000/1999)

32/100

Tính

\(\left[ {\left( {1 + \frac{{1999}}{1}} \right)\left( {1 + \frac{{1999}}{2}} \right)...\left( {1 + \frac{{1999}}{{1000}}} \right)} \right]:\left[ {\left( {1 + \frac{{1000}}{1}} \right)\left( {1 + \frac{{1000}}{2}} \right)\left( {1 + \frac{{1000}}{{1999}}} \right)} \right]\)

Nhập câu trả lời ở đây

0/3000 ký tự

Giải thích

Lời giải:

\(\begin{array}{l}\left[ {\left( {1 + \frac{{1999}}{1}} \right)\left( {1 + \frac{{1999}}{2}} \right)...\left( {1 + \frac{{1999}}{{1000}}} \right)} \right]:\left[ {\left( {1 + \frac{{1000}}{1}} \right)\left( {1 + \frac{{1000}}{2}} \right)\left( {1 + \frac{{1000}}{{1999}}} \right)} \right]\\ = (\frac{{2000}}{1}.\frac{{2001}}{2}...\frac{{2999}}{{1000}}):(\frac{{1001}}{1}.\frac{{1002}}{2}...\frac{{1999}}{{2999}})\\ = \frac{{2000.2001...2999}}{{1.2...1000}}.\frac{{1.2...1999}}{{1001.1002...2999}}\\ = 1\end{array}\)