Đáp án đúng: B
Giải chi tiết:Hướng dẫn giải chi tiết
\(A = \dfrac{1}{{1.3}} + \dfrac{1}{{2.4}} + \dfrac{1}{{3.5}} + ... + \dfrac{1}{{7.9}} + \dfrac{1}{{8.10}}\)
Áp dụng tính chất giao hoán và kết hợp của phép cộng ta có:
\(\begin{array}{l}A = \dfrac{1}{{1.3}} + \dfrac{1}{{2.4}} + \dfrac{1}{{3.5}} + ... + \dfrac{1}{{7.9}} + \dfrac{1}{{8.10}}\\\,\,\,\,\, = \left( {\dfrac{1}{{1.3}} + \dfrac{1}{{3.5}} + ... + \dfrac{1}{{7.9}}} \right) + \left( {\dfrac{1}{{2.4}} + \dfrac{1}{{4.6}} + ... + \frac{1}{{8.10}}} \right)\\\,\,\,\,\, = \dfrac{1}{2}\left( {\dfrac{1}{1} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{5} + ... + \dfrac{1}{7} - \dfrac{1}{9}} \right) + \dfrac{1}{2}\left( {\dfrac{1}{2} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{6} + ... + \dfrac{1}{8} - \dfrac{1}{{10}}} \right)\\\,\,\,\,\, = \dfrac{1}{2}\left( {1 - \dfrac{1}{9}} \right) + \dfrac{1}{2}\left( {\dfrac{1}{2} - \dfrac{1}{{10}}} \right)\\\,\,\,\,\, = \dfrac{1}{2}\left( {1 - \dfrac{1}{9} + \dfrac{1}{2} - \dfrac{1}{{10}}} \right)\\\,\,\,\,\, = \dfrac{1}{2}.\dfrac{{90 + 45 - 10 - 9}}{{90}}\\\,\,\,\,\, = \dfrac{1}{2}.\dfrac{{116}}{{90}} = \dfrac{{29}}{{45}}\end{array}\)
