Đáp án đúng: C
Giải chi tiết:\(\left( {\frac{1}{{1.2.3}} + \frac{1}{{2.3.4}} + \ldots + \frac{1}{{8.9.10}}} \right) \cdot x = \frac{{23}}{{45}}\)
\(\begin{array}{l}\left( {\frac{1}{{1.2.3}} + \frac{1}{{2.3.4}} + \ldots + \frac{1}{{8.9.10}}} \right) \cdot x = \frac{{23}}{{45}}\\2 \cdot \left( {\frac{1}{{1.2.3}} + \frac{1}{{2.3.4}} + \ldots + \frac{1}{{8.9.10}}} \right) \cdot x = \frac{{23}}{{45}} \cdot 2\\\left( {\frac{2}{{1.2.3}} + \frac{2}{{2.3.4}} + \ldots + \frac{2}{{8.9.10}}} \right) \cdot x = \frac{{46}}{{45}}\\\left( {\frac{{3 - 1}}{{1.2.3}} + \frac{{4 - 2}}{{2.3.4}} + \ldots + \frac{{10 - 8}}{{8.9.10}}} \right) \cdot x = \frac{{46}}{{45}}\\\left[ {\left( {\frac{1}{{1.2}} - \frac{1}{{2.3}}} \right) + \left( {\frac{1}{{2.3}} - \frac{1}{{3.4}}} \right) + \ldots + \left( {\frac{1}{{8.9}} - \frac{1}{{9.10}}} \right)} \right] \cdot x = \frac{{46}}{{45}}\\\left( {\frac{1}{{1.2}} - \frac{1}{{2.3}} + \frac{1}{{2.3}} - \frac{1}{{3.4}} + \ldots + \frac{1}{{8.9}} - \frac{1}{{9.10}}} \right) \cdot x = \frac{{46}}{{45}}\\\left( {\frac{1}{{1.2}} - \frac{1}{{9.10}}} \right) \cdot x = \frac{{46}}{{45}}\\\left( {\frac{1}{2} - \frac{1}{{90}}} \right) \cdot x = \frac{{46}}{{45}}\\\left( {\frac{{45}}{{90}} - \frac{1}{{90}}} \right) \cdot x = \frac{{46}}{{45}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{44}}{{90}} \cdot x = \frac{{46}}{{45}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \frac{{46}}{{45}}:\frac{{44}}{{90}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \frac{{92}}{{44}} = \frac{{23}}{{11}}\end{array}\)
Vậy \(x = \frac{{23}}{{11}}\).
Chọn C.
