Đáp án đúng: C
Giải chi tiết:\(\frac{1}{{5.8}} + \frac{1}{{8.11}} + \frac{1}{{11.14}} + \ldots + \frac{1}{{x.\left( {x + 3} \right)}} = \frac{{101}}{{1540}}\)
\(\begin{array}{l}\frac{1}{{5.8}} + \frac{1}{{8.11}} + \frac{1}{{11.14}} + \ldots + \frac{1}{{x.\left( {x + 3} \right)}} = \frac{{101}}{{1540}}\\\frac{3}{{5.8}} + \frac{3}{{8.11}} + \frac{3}{{11.14}} + \ldots + \frac{3}{{x.\left( {x + 3} \right)}} = \frac{{303}}{{1540}}\\\left( {\frac{1}{5} - \frac{1}{8}} \right) + \left( {\frac{1}{8} - \frac{1}{{11}}} \right) + \left( {\frac{1}{{11}} - \frac{1}{{14}}} \right) + \ldots + \left( {\frac{1}{x} - \frac{1}{{x + 3}}} \right) = \frac{{303}}{{1540}}\\\frac{1}{5} - \frac{1}{8} + \frac{1}{8} - \frac{1}{{11}} + \frac{1}{{11}} - \frac{1}{{14}} + \ldots + \frac{1}{x} - \frac{1}{{x + 3}} = \frac{{303}}{{1540}}\end{array}\)
\(\begin{array}{l}\frac{1}{5} - \frac{1}{{x + 3}} = \frac{{303}}{{1540}}\\\,\,\,\,\,\,\,\,\,\frac{1}{{x + 3}} = \frac{1}{5} - \frac{{303}}{{1540}}\\\,\,\,\,\,\,\,\,\,\frac{1}{{x + 3}} = \frac{{308}}{{1540}} - \frac{{303}}{{1540}}\\\,\,\,\,\,\,\,\,\frac{1}{{x + 3}} = \frac{5}{{1540}}\\\,\,\,\,\,\,\,\,\frac{1}{{x + 3}} = \frac{1}{{308}}\\ \Rightarrow x + 3 = 308\\ \Rightarrow x = 305\end{array}\)
Vậy \(x = 305\).
Chọn C.
