Tính \(\frac{{\left( {1 + \frac{{1999}}{1}} \right)\left( {1 + \frac{{1999}}{2}} \right) \ldots \left( {1 + \frac{{1999}}{{1000}}} \right)}}{{\left( {1 + \frac{{1000}}{1}} \right)\left( {1 + \frac{{1000}}{2}} \right) \ldots \left( {1 + \frac{{1000}}{{1999}}} \right)}}\) ta được
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Lời giải:
Báo sai\(\begin{array}{l} \frac{{\left( {1 + \frac{{1999}}{1}} \right)\left( {1 + \frac{{1999}}{2}} \right) \ldots \left( {1 + \frac{{1999}}{{1000}}} \right)}}{{\left( {1 + \frac{{1000}}{1}} \right)\left( {1 + \frac{{1000}}{2}} \right) \ldots \left( {1 + \frac{{1000}}{{1999}}} \right)}} = \frac{{\frac{{2000}}{1} \cdot \frac{{2001}}{2} \cdot \frac{{2002}}{3} \ldots \frac{{2999}}{{1000}}}}{{\frac{{1001}}{1} \cdot \frac{{1002}}{2} \cdot \frac{{1003}}{3} \ldots .\frac{{2999}}{{1999}}}}\\ A = \left( {\frac{{2000}}{1} \cdot \frac{{2001}}{2} \cdot \frac{{2002}}{3} \ldots \frac{{2999}}{{1000}}} \right):\left( {\frac{{1001}}{1} \cdot \frac{{1002}}{2} \cdot \frac{{1003}}{3} \ldots .\frac{{2999}}{{1999}}} \right)\\ A = \left( {\frac{{2000.2001.2002 \ldots 2999}}{{1.2.3.4 \ldots 1000}}} \right) \cdot \left( {\frac{{1.2.3 \ldots 1999}}{{1001.1002 \ldots 2999}}} \right) = \frac{{1001.1002 \ldots .1999}}{{1001.1002 \ldots 1999}} = 1 \end{array}\)