Đề bài là $\frac{3}{7}x(\frac{3535}{1212} + \frac{3535}{2020} + \frac{3535}{3030} + \frac{3535}{4242} + \frac{3535}{5656}) = \frac{25}{24}$ phải không bạn?
$\frac{3}{7} . x . (\frac{3535}{1212} + \frac{3535}{2020} + \frac{3535}{3030} + \frac{3535}{4242} + \frac{3535}{5656}) = \frac{25}{24}$
\Leftrightarrow $\frac{3}{7} . x . \frac{3535}{101} . (\frac{1}{12} + \frac{!}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56}) = \frac{25}{24}$
\Leftrightarrow $15x . (\frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6} + \frac{1}{6} - \frac{1}{7} + \frac{1}{7} - \frac{1}{8} = \frac{25}{24}$
\Leftrightarrow $15x . (\frac{1}{3} - \frac{1}{8}) = \frac{25}{24}$
\Leftrightarrow $15x . \frac{5}{24} = \frac{25}{24}$
\Leftrightarrow $15x = 5$
\Leftrightarrow $x = \frac{1}{3}$
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