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Tính bằng cách hợp lí :
a) \({{ – 2} \over 5}.{5 \over 8} + {5 \over 8}.{3 \over 5}\)
b) \({{ – 3} \over 5}.{5 \over 7} + {{ – 3} \over 5}.{3 \over 7} + {{ – 3} \over 5}.{6 \over 7}\)
c) \({4 \over {19}}.{{ – 5} \over 7} + {{ – 5} \over 7}.{{15} \over {19}} + 1{4 \over 7}\)
d) \(7{5 \over 9} – \left( {2{3 \over 4} + 3{5 \over 9}} \right)\)
e) \(7{5 \over {11}} – \left( {2{3 \over 7} + 3{5 \over {11}}} \right)\)
f) \(\left( {{{ – 4} \over 5} + {4 \over 3}} \right) + \left( {{{ – 5} \over 4} + {{14} \over 5}} \right) – {7 \over 3}\).
\(\eqalign{ & a){{ – 2} \over 5}.{5 \over 8} + {5 \over 8}.{3 \over 5} = {5 \over 8}.\left( {{{ – 2} \over 5} + {3 \over 5}} \right) = {5 \over 8}.{1 \over 5} = {1 \over 8}. \cr & b){{ – 3} \over 5}.{5 \over 7} + {{ – 3} \over 5}.{3 \over 7} + {{ – 3} \over 5}.{6 \over 7} = {{ – 3} \over 5}.\left( {{5 \over 7} + {3 \over 7} + {6 \over 7}} \right) \cr & = {{ – 3} \over 5}.{{14} \over 7} = {{ – 3} \over 5}.2 = {{ – 6} \over 5} = – 1{1 \over 5} \cr & c){4 \over {19}}.{{ – 5} \over 7} + {{ – 5} \over 7}.{{15} \over {19}} + 1{4 \over 7} = {{ – 5} \over 7}.\left( {{4 \over {19}} + {{15} \over {19}}} \right) + {{11} \over 7} \cr & = {{ – 5} \over 7}.{{19} \over {19}} + {{11} \over 7} = {{ – 5} \over 7} + {{11} \over 7} = {6 \over 7} \cr & d)7{5 \over 9} – \left( {2{3 \over 4} + 3{5 \over 9}} \right) = \left( {7{5 \over 9} – 3{5 \over 9}} \right) – 2{3 \over 4} \cr & = 4 – 2{3 \over 4} = 3{4 \over 4} – 2{3 \over 4} = 1{1 \over 4} \cr & e)7{5 \over {11}} – \left( {2{3 \over 7} + 3{5 \over {11}}} \right) = \left( {7{5 \over {11}} – 3{5 \over {11}}} \right) – 2{3 \over 7} \cr & = 4 – 2{3 \over 7} = 3{7 \over 7} – 2{3 \over 7} = 1{4 \over 7} \cr & f)\left( {{{ – 4} \over 5} + {4 \over 3}} \right) + \left( {{{ – 5} \over 4} + {{14} \over 5}} \right) – {7 \over 3} = \left( {{{ – 4} \over 5} + {{14} \over 5}} \right) + \left( {{4 \over 3} – {7 \over 3}} \right) + {{ – 5} \over 4} \cr & = {{10} \over 5} + {{ – 3} \over 3} + {{ – 5} \over 4} = 2 + ( – 1) + {{ – 5} \over 4} = 1 + {{ – 5} \over 4} = {4 \over 4} – {5 \over 4} = {{ – 1} \over 4}. \cr} \)