Tính
a) \( – 0,375 + \dfrac{1}{4} – \left( { – \dfrac{1}{5}} \right)\)
b) \(\dfrac{2}{3} + \left( {\dfrac{{ – 5}}{{12}}} \right) + \left( {\dfrac{{ – 7}}{{15}}} \right) + ( – 0,2)\)
c) \(0,275 + \left( {\dfrac{{ – 8}}{{17}}} \right) + \dfrac{{29}}{{40}} + \left( {\dfrac{{ – 9}}{{17}}} \right) – 1\dfrac{1}{3}\)
d) \(\left( { – 5} \right).\left( {\dfrac{{ – 34}}{{21}}} \right).\left( {\dfrac{{ – 7}}{{10}}} \right).\left( { – \dfrac{3}{{17}}} \right)\)
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e) \(\left( {\dfrac{{13}}{{18}}:\dfrac{{26}}{9}} \right).\dfrac{4}{7}\)
f) \(\left[ {\left( { – \dfrac{5}{{33}}} \right):\dfrac{2}{{11}}} \right].0,15\)
Ta đổi những số hữu tỉ về dạng phân số sau đó áp dụng các qui tắc nhân chia phân số
\(\begin{array}{l}a) – 0,375 + \dfrac{1}{4} – \left( { – \dfrac{1}{5}} \right) = \dfrac{{ – 3}}{8} + \dfrac{1}{4} + \dfrac{1}{5}\\ = \dfrac{{ – 15}}{{40}} + \dfrac{{10}}{{40}} + \dfrac{8}{{40}} = \dfrac{3}{{40}}\\b)\dfrac{2}{3} + \left( {\dfrac{{ – 5}}{{12}}} \right) + \left( {\dfrac{{ – 7}}{{15}}} \right) + ( – 0,2) = \dfrac{2}{3} – \dfrac{5}{{12}} – {\dfrac{7}{{15}} – \dfrac{1}{5}}\\ = \left( {\dfrac{2}{3} – \dfrac{5}{{12}}} \right) – \left( {\dfrac{7}{{15}} + \dfrac{1}{5}} \right) = \left( {\dfrac{8}{{12}} – \dfrac{5}{{12}}} \right) – \left( {\dfrac{7}{{15}} + \dfrac{3}{{15}}} \right)\\ = \dfrac{3}{{12}} – \dfrac{{10}}{{15}}= \dfrac{1}{{4}} – \dfrac{{2}}{{3}} = \dfrac{3}{{12}} – \dfrac{8}{{12}} = \dfrac{{ – 5}}{{12}}\\c)0,275 + \left( {\dfrac{{ – 8}}{{17}}} \right) + \dfrac{{29}}{{40}} + \left( {\dfrac{{ – 9}}{{17}}} \right) – 1\dfrac{1}{3}\\ = \dfrac{{11}}{{40}} + \left( {\dfrac{{ – 8}}{{17}}} \right) + \dfrac{{29}}{{40}} + \left( {\dfrac{{ – 9}}{{17}}} \right) – \dfrac{4}{3}\\ = \left( {\dfrac{{11}}{{40}} + \dfrac{{29}}{{40}}} \right) + \left[ {\left( {\dfrac{{ – 8}}{{17}}} \right) + \left( {\dfrac{{ – 9}}{{17}}} \right)} \right] – \dfrac{4}{3}\\ =\dfrac{40}{40}+\dfrac{-17}{17}-\dfrac{4}{3}= 1 + ( – 1) – \dfrac{4}{3} = – \dfrac{4}{3}\\d)\left( { – 5} \right).\left( {\dfrac{{ – 34}}{{21}}} \right).\left( {\dfrac{{ – 7}}{{10}}} \right).\left( { – \dfrac{3}{{17}}} \right)\\ = 5.\dfrac{{34}}{{21}}.\dfrac{7}{{10}}.\dfrac{3}{{17}} = \dfrac{{5.34.7.3}}{{21.10.17}} = \dfrac{{5.17.2.7.3}}{{7.3.5.2.17}} = 1\\e)\left( {\dfrac{{13}}{{18}}:\dfrac{{26}}{9}} \right).\dfrac{4}{7} = \dfrac{{13}}{{18}}.\dfrac{9}{{26}}.\dfrac{4}{7} = \dfrac{{13.9.4}}{{18.26.7}} = \dfrac{{13.9.2.2}}{{2.9.13.2.7}} = \dfrac{1}{7}\\f)\left[ {\left( { – \dfrac{5}{{33}}} \right):\dfrac{2}{{11}}} \right].0,15 = \dfrac{{ – 5}}{{33}}.\dfrac{{11}}{2}.\dfrac{3}{{20}} = – \dfrac{{5.11.3}}{{3.11.2.4.5}} = \dfrac{{ – 1}}{8}\end{array}\)