Tìm x biết \(\left( {\frac{1}{{1.2}} + \frac{1}{{3.4}} + \ldots + \frac{1}{{99.100}}} \right)x = \frac{{2012}}{{51}} + \frac{{2012}}{{52}} + \ldots + \frac{{2012}}{{100}}\)
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Lời giải:
Báo sai\(\begin{array}{l} \left( {\frac{1}{{1.2}} + \frac{1}{{3.4}} + \ldots + \frac{1}{{99.100}}} \right)x = \frac{{2012}}{{51}} + \frac{{2012}}{{52}} + \ldots + \frac{{2012}}{{100}}\\ Ta\,có:\\ \frac{1}{{1.2}} + \frac{1}{{3.4}} + \ldots + \frac{1}{{99.100}} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{{99}} - \frac{1}{{100}}\\ = \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{{99}} + \frac{1}{{100}}} \right) - 2\left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \ldots + \frac{1}{{100}}} \right)\\ = \left( {\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{{100}}} \right) - \left( {\frac{1}{1} + \frac{1}{2} + \ldots + \frac{1}{{50}}} \right) = \frac{1}{{51}} + \frac{1}{{52}} + \frac{1}{{53}} + \ldots + \frac{1}{{100}}\\ Khi\,đó\\ \left( {\frac{1}{{51}} + \frac{1}{{52}} + \ldots + \frac{1}{{100}}} \right)x = 2012\left( {\frac{1}{{51}} + \frac{1}{{52}} + \frac{1}{{53}} + \ldots + \frac{1}{{100}}} \right) = > x = 2012\\ \end{array}\)