A=(1/2+1/3+.....+1/2021).(1+1/2+1/3+.....+1/2020)-(1+1/2+1/3+........+1/2021).(1/2+1/3+.....1/2020)
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1 câu trả lời 317
\[\begin{array}{l}
A = \left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2021}}} \right).\left( {1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2020}}} \right) - \left( {1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2021}}} \right)\left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2020}}} \right)\\
= \left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2021}}} \right) + \left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2021}}} \right).\left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2020}}} \right) - \left[ {\left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2020}}} \right) + \left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2021}}} \right)\left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2020}}} \right)} \right]\\
= \left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2021}}} \right) + \left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2021}}} \right).\left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2020}}} \right) - \left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2020}}} \right) - \left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2021}}} \right)\left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2020}}} \right)\\
= \left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2021}} - \frac{1}{2} - \frac{1}{3} - ... - \frac{1}{{2020}}} \right) + \left[ {\left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2021}}} \right).\left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2020}}} \right) - \left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2021}}} \right).\left( {\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{2020}}} \right)} \right]\\
= \frac{1}{{2021}}
\end{array}\]
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