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) \(B = {5 \over {2.1}} + {4 \over {1.11}} + {3 \over {11.2}} + {1 \over {2.15}} + {{13} \over {15.4}}\).
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\(\eqalign{ & a)A = {{ – 1} \over {20}} + {{ – 1} \over {30}} + {{ – 1} \over {42}} + {{ – 1} \over {56}} + {{ – 1} \over {72}} + {{ – 1} \over {90}} \cr & = {{ – 1} \over {4.5}} + {{ – 1} \over {5.6}} + {{ – 1} \over {6.7}} + {{ – 1} \over {7.8}} + {{ – 1} \over {8.9}} + {{ – 1} \over {9.10}} \cr & = {{4 – 5} \over {4.5}} + {{5 – 6} \over {5.6}} + {{6 – 7} \over {6.7}} + {{7 – 8} \over {7.8}} + {{8 – 9} \over {8.9}} + {{9 – 10} \over {9.10}} \cr & = {1 \over 5} – {1 \over 4} + {1 \over 6} – {1 \over 5} + {1 \over 7} – {1 \over 6} + {1 \over 8} – {1 \over 7} + {1 \over 9} – {1 \over 8} + {1 \over {10}} – {1 \over 9} \cr & = {{ – 1} \over 4} + {1 \over {10}} = {{ – 5} \over {20}} + {2 \over {20}} = {{ – 3} \over {20}}. \cr & b)B = {5 \over {2.1}} + {4 \over {1.11}} + {3 \over {11.2}} + {1 \over {2.15}} + {{13} \over {15.4}} \cr & = 7.\left( {{5 \over {2.7}} + {4 \over {7.11}} + {3 \over {11.14}} + {1 \over {14.15}} + {{13} \over {15.28}}} \right) \cr & = 7.\left( {{{7 – 2} \over {2.7}} + {{11 – 7} \over {7.11}} + {{14 – 11} \over {11.14}} + {{15 – 14} \over {14.15}} + {{28 – 15} \over {15.28}}} \right) \cr & = 7\left( {{1 \over 2} – {1 \over 7} + {1 \over 7} – {1 \over {11}} + {1 \over {11}} – {1 \over {14}} + {1 \over {14}} – {1 \over {15}} + {1 \over {15}} – {1 \over {28}}} \right) \cr & = 7.\left( {{1 \over 2} – {1 \over {28}}} \right) = 7.\left( {{{14} \over {28}} – {1 \over {28}}} \right) = 7{3 \over {28}} = {{13} \over 4} = 3{1 \over 4}. \cr} \)