Cho \({\rm{M}} = \frac{{{2^{2018}}}}{{{2^{2018}} + {3^{2019}}}} + \frac{{{3^{2019}}}}{{{3^{2019}} + {5^{2020}}}} + \frac{{{5^{2020}}}}{{{5^{2020}} + {2^{2019}}}};{\rm{N}} = \frac{1}{{1.2}} + \frac{1}{{3.4}} + \frac{1}{{5.6}} + \ldots \ldots \ldots + \frac{1}{{2019.2020}}\). Khẳng định nào sau đây đúng?
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Lời giải:
Báo sai\(\begin{array}{l} {\rm{M}} = \frac{{{2^{2018}}}}{{{2^{2018}} + {3^{2019}}}} + \frac{{{3^{2019}}}}{{{3^{2019}} + {5^{2020}}}} + \frac{{{5^{2020}}}}{{{5^{2020}} + {2^{2019}}}}\\ > \frac{{{2^{2018}}}}{{{2^{2018}} + {3^{2019}} + {5^{2020}}}} + \frac{{{3^{2019}}}}{{{3^{2019}} + {5^{2020}} + {2^{2018}}}} + \frac{{{5^{2020}}}}{{{5^{2020}} + {2^{2019}} + {3^{2019}}}} = \frac{{{2^{2018}} + {3^{2019}} + {5^{2020}}}}{{{2^{2018}} + {3^{2019}} + {5^{2020}}}} = 1\\ {\rm{N}} = \frac{1}{{1.2}} + \frac{1}{{3.4}} + \frac{1}{{5.6}} + \ldots \ldots \ldots + \frac{1}{{2019.2020}}\\ < \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + \frac{1}{{4.5}} + \frac{1}{{5.6}} + \ldots \ldots \ldots + \frac{1}{{2019.2020}} = 1 - \frac{1}{{2020}} < 1\\ \Rightarrow M > 1 > N \end{array}\)