Tìm x biết \((x - 20)\frac{{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{{200}}}}{{\frac{1}{{199}} + \frac{2}{{198}} + \ldots + \frac{{199}}{1}}} = \frac{1}{{2000}}\)
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Lời giải:
Báo sai\(\begin{array}{l} (x - 20)\frac{{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{{200}}}}{{\frac{1}{{199}} + \frac{2}{{198}} + \ldots + \frac{{199}}{1}}} = \frac{1}{{2000}}\\ A = \frac{{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{{200}}}}{{\frac{1}{{199}} + \frac{2}{{198}} + \ldots + \frac{{199}}{1}}}\\ Mẫu\,của\,A \,là: \left( {\frac{1}{{199}} + 1} \right) + \left( {\frac{2}{{198}} + 1} \right) + \ldots + \left( {\frac{{198}}{2} + 1} \right) + 1 = \frac{{200}}{{199}} + \frac{{200}}{{198}} + \ldots + \frac{{200}}{2} + \frac{{200}}{{200}}\\ \Rightarrow A = \frac{{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{{200}}}}{{200\left( {\frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{{200}}} \right)}} = \frac{1}{{200}}\\ Ta\,có\\ (x - 20) \cdot \frac{1}{{200}} = \frac{1}{{2000}} \Leftrightarrow x - 20 = \frac{1}{{10}} \Rightarrow x = \frac{1}{{10}} - 20 = \frac{{ - 199}}{{10}} \end{array}\)