Bài 1: Tính:
a) \(4 – 1{2 \over 5} – {8 \over 3}\)
b) \( – 1 + {1 \over 3} – {1 \over 9} – {1 \over {81}}\)
c) \({4 \over 5} – \left) { – {2 \over 7}} \right) – {7 \over {10}}\)
d) \({3 \over 4} – \left[ {{3 \over 4} – \left( {{2 \over 3} + {5 \over 6}} \right)} \right]\)
Bài 2: Tính bằng cách hợp lí:
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\(S = {3 \over 4} – 0,25 – \left[ {{7 \over 3} + \left( { – {9 \over 2}} \right)} \right] – {5 \over 6}.\)
Bài 1: a) \(4 – 1{2 \over 5} – {8 \over 3} = 4 – {7 \over 5} – {8 \over 3} \)
\(\;= {{60} \over {15}} – {{21} \over {15}} – {{40} \over {15}} = {{60 – 21 – 40} \over {15}} = {{ – 1} \over {15}}.\)
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b) \( – 1 + {1 \over 3} – {1 \over 9} – {1 \over {81}}\)
\(\;\;= {{ – 81} \over {81}} + {{27} \over {81}} – {9 \over {81}} – {1 \over {81}}\)
\(\;\;={{ – 81 + 27 – 9 – 1} \over {81}} = {{ – 64} \over {81}}.\)
c) \({4 \over 5} – \left( { – {2 \over 7}} \right) – {7 \over {10}} = {4 \over 5} + {2 \over 7} – {7 \over {10}} \)
\(= {{56} \over {70}} + {{20} \over {70}} – {{49} \over {70}} = {{56 + 20 – 49} \over {70}} = {{27} \over {70}}\)
d) \({3 \over 4} – \left[ {{3 \over 4} – \left( {{2 \over 3} + {5 \over 6}} \right)} \right] = {3 \over 4} – \left( {{3 \over 4} – {9 \over 6}} \right)\)
\( = {3 \over 4} – {3 \over 4} + {9 \over 6} = {9 \over 6} = {3 \over 2}.\)
Bài 2: \(S = {3 \over 4} – 0,25 – \left[ {{7 \over 3} + \left( { – {9 \over 2}} \right)} \right] – {5 \over 6}\)
\(\eqalign{ & = {3 \over 4} – {1 \over 4} – {7 \over 3} + {9 \over 2} – {5 \over 6} \cr & = \left( {{3 \over 4} – {1 \over 4}} \right) – \left( {{7 \over 3} – {9 \over 2} + {5 \over 6}} \right) \cr & = {2 \over 4} – \left( {{{14} \over 6} – {{27} \over 6} + {5 \over 6}} \right) \cr & = {1 \over 2} – \left( { – {8 \over 6}} \right) \cr & = {1 \over 2} + {4 \over 3} = {{3 + 8} \over 6} = {{11} \over 6}. \cr} \)
