Không quy đồng, hãy tính các tổng sau một cách hợp lí :
a) \(A = {{ - 1} \over {20}} + {{ - 1} \over {30}} + {{ - 1} \over {42}} + {{ - 1} \over {56}} + {{ - 1} \over {72}} + {{ - 1} \over {90}}\)
b) \(B = {5 \over {2.1}} + {4 \over {1.11}} + {3 \over {11.2}} + {1 \over {2.15}} + {{13} \over {15.4}}\).
\(\eqalign{ & a)A = {{ - 1} \over {20}} + {{ - 1} \over {30}} + {{ - 1} \over {42}} + {{ - 1} \over {56}} + {{ - 1} \over {72}} + {{ - 1} \over {90}} \cr & = {{ - 1} \over {4.5}} + {{ - 1} \over {5.6}} + {{ - 1} \over {6.7}} + {{ - 1} \over {7.8}} + {{ - 1} \over {8.9}} + {{ - 1} \over {9.10}} \cr & = {{4 - 5} \over {4.5}} + {{5 - 6} \over {5.6}} + {{6 - 7} \over {6.7}} + {{7 - 8} \over {7.8}} + {{8 - 9} \over {8.9}} + {{9 - 10} \over {9.10}} \cr & = {1 \over 5} - {1 \over 4} + {1 \over 6} - {1 \over 5} + {1 \over 7} - {1 \over 6} + {1 \over 8} - {1 \over 7} + {1 \over 9} - {1 \over 8} + {1 \over {10}} - {1 \over 9} \cr & = {{ - 1} \over 4} + {1 \over {10}} = {{ - 5} \over {20}} + {2 \over {20}} = {{ - 3} \over {20}}. \cr & b)B = {5 \over {2.1}} + {4 \over {1.11}} + {3 \over {11.2}} + {1 \over {2.15}} + {{13} \over {15.4}} \cr & = 7.\left( {{5 \over {2.7}} + {4 \over {7.11}} + {3 \over {11.14}} + {1 \over {14.15}} + {{13} \over {15.28}}} \right) \cr & = 7.\left( {{{7 - 2} \over {2.7}} + {{11 - 7} \over {7.11}} + {{14 - 11} \over {11.14}} + {{15 - 14} \over {14.15}} + {{28 - 15} \over {15.28}}} \right) \cr & = 7\left( {{1 \over 2} - {1 \over 7} + {1 \over 7} - {1 \over {11}} + {1 \over {11}} - {1 \over {14}} + {1 \over {14}} - {1 \over {15}} + {1 \over {15}} - {1 \over {28}}} \right) \cr & = 7.\left( {{1 \over 2} - {1 \over {28}}} \right) = 7.\left( {{{14} \over {28}} - {1 \over {28}}} \right) = 7{3 \over {28}} = {{13} \over 4} = 3{1 \over 4}. \cr} \)