Tìm x biết \(\frac{1}{{3.5}} + \frac{1}{{5.7}} + \frac{1}{{7.9}} + \ldots + \frac{1}{{(2x + 1)(2x + 3)}} = \frac{{15}}{{93}}\)
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Lời giải:
Báo sai\(\begin{array}{l} \frac{1}{{3.5}} + \frac{1}{{5.7}} + \frac{1}{{7.9}} + \ldots + \frac{1}{{(2x + 1)(2x + 3)}} = \frac{{15}}{{93}}\\ \Leftrightarrow \left( {\frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \ldots . + \frac{1}{{2x + 1}} - \frac{1}{{2x + 3}}} \right) \cdot \frac{1}{2} = \frac{{15}}{{93}}\\ \Leftrightarrow \left( {\frac{1}{3} - \frac{1}{{2x + 3}}} \right) \cdot \frac{1}{2} = \frac{{15}}{{93}}\\ \Leftrightarrow \frac{1}{3} - \frac{1}{{2x + 3}} = \frac{{30}}{{93}} \Leftrightarrow \frac{1}{{2x + 3}} = \frac{1}{3} - \frac{{30}}{{93}}\\ \Leftrightarrow \frac{1}{{2x + 3}} = \frac{1}{{93}} \Leftrightarrow 2x + 3 = 93 \Leftrightarrow x = 45 \end{array}\)