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a)\(\left( { – 0,5} \right) – \left( { – 1 + \dfrac{2}{3}} \right):1,5 + \left( {\dfrac{{ – 1}}{4}} \right)\)
b)\(\left[ {\left( {\dfrac{{ – 7}}{8}} \right):\dfrac{{21}}{{16}}} \right] – \dfrac{5}{3}.\left( {\dfrac{1}{3} – \dfrac{7}{{10}}} \right)\)
c)\({\left[ {\left( {\dfrac{{ – 2}}{3}} \right) + \dfrac{3}{4}} \right]^2}.\dfrac{{12}}{5} – \dfrac{1}{5}\)
d)\({\left( {\dfrac{1}{{25}} – 0,4} \right)^2}:\dfrac{9}{{125}} – \left[ {\left( {1\dfrac{1}{3} – \dfrac{2}{5}} \right).\dfrac{3}{7}} \right]\)
e)\(\left\{ {3\dfrac{{17}}{{18}}.\left[ {\dfrac{5}{2} – \left( {\dfrac{1}{3} + \dfrac{2}{9}} \right)} \right]} \right\}:{\left[ {\left( {\dfrac{{ – 1}}{2}} \right) + 0,25} \right]^2}\)
Áp dụng quy tắc bỏ ngoặc rồi tính toán, nếu có lũy thừa hay số thập phân thì ta viết chúng dưới dạng phân số để thuận lợi trong tính toán
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a)\(\left( { – 0,5} \right) – \left( { – 1 + \dfrac{2}{3}} \right):1,5 + \left( {\dfrac{{ – 1}}{4}} \right)\)
\(\begin{array}{l} = \left( {\dfrac{{ – 1}}{2}} \right) – \left( {\dfrac{{ – 3}}{3} + \dfrac{2}{3}} \right):\dfrac{3}{2} + \left( {\dfrac{{ – 1}}{4}} \right)\\ = \left( {\dfrac{{ – 1}}{2}} \right) – \left( {\dfrac{{ – 1}}{3}} \right).\dfrac{2}{3} + \left( {\dfrac{{ – 1}}{4}} \right)\\ = \left( {\dfrac{{ – 1}}{2}} \right) + \dfrac{2}{9} + \left( {\dfrac{{ – 1}}{4}} \right)\\ = \left( {\dfrac{{ – 18}}{{36}}} \right) + \dfrac{8}{{36}} + \left( {\dfrac{{ – 9}}{{36}}} \right) = \dfrac{{ – 19}}{{36}}\end{array}\)
b)\(\left[ {\left( {\dfrac{{ – 7}}{8}} \right):\dfrac{{21}}{{16}}} \right] – \dfrac{5}{3}.\left( {\dfrac{1}{3} – \dfrac{7}{{10}}} \right)\)
\(\begin{array}{l} = \left[ {\left( {\dfrac{{ – 7}}{8}} \right).\dfrac{{16}}{{21}}} \right] – \dfrac{5}{3}.\left( {\dfrac{{10}}{{30}} – \dfrac{{21}}{{30}}} \right)\\ = \dfrac{{\left( { – 7} \right).16}}{{8.21}} – \dfrac{5}{3}.\left( {\dfrac{{ – 11}}{{30}}} \right)\end{array}\)
\(\begin{array}{l} = – \dfrac{{7.8.2}}{{8.7.3}} + \dfrac{{5.11}}{{3.5.6}}\\ = \dfrac{{ – 2}}{3} + \dfrac{{11}}{{18}} = \dfrac{{ – 1}}{{18}}\end{array}\)
c)\({\left[ {\left( {\dfrac{{ – 2}}{3}} \right) + \dfrac{3}{4}} \right]^2}.\dfrac{{12}}{5} – \dfrac{1}{5}\) \( = {\left[ {\left( {\dfrac{{ – 8}}{{12}}} \right) + \dfrac{9}{{12}}} \right]^2}.\dfrac{{12}}{5} – \dfrac{1}{5} = {\left( {\dfrac{1}{{12}}} \right)^2}.\dfrac{{12}}{5} – \dfrac{1}{5}\\ = \dfrac{1}{{{{12}^2}}}.\dfrac{{12}}{5} – \dfrac{1}{5} = \dfrac{1}{{60}} – \dfrac{1}{5}= \dfrac{1}{{60}} – \dfrac{12}{60} = \dfrac{{ – 11}}{{60}}\)
d)\({\left( {\dfrac{1}{{25}} – 0,4} \right)^2}:\dfrac{9}{{125}} – \left[ {\left( {1\dfrac{1}{3} – \dfrac{2}{5}} \right).\dfrac{3}{7}} \right]\)
\(\begin{array}{l} = {\left( {\dfrac{1}{{25}} – \dfrac{2}{5}} \right)^2}.\dfrac{{125}}{9} – \left[ {\left( {\dfrac{4}{3} – \dfrac{2}{5}} \right).\dfrac{3}{7}} \right]\\ = {\left( {\dfrac{{ – 9}}{{25}}} \right)^2}.\dfrac{{125}}{9} – \left( {\dfrac{{14}}{{15}}.\dfrac{3}{7}} \right)\\ = \dfrac{{{9^2}}}{{{{25}^2}}}.\dfrac{{125}}{9} – \dfrac{2}{5}\\ = \dfrac{{{{\left( {{3^2}} \right)}^2}}}{{{{\left( {{5^2}} \right)}^2}}}.\dfrac{{{5^3}}}{{{3^2}}} – \dfrac{2}{5} = \dfrac{{{3^2}}}{5} – \dfrac{2}{5} = \dfrac{9}{5} – \dfrac{2}{5} = \dfrac{7}{5}\end{array}\)
e)\(\left\{ {3\dfrac{{17}}{{18}}.\left[ {\dfrac{5}{2} – \left( {\dfrac{1}{3} + \dfrac{2}{9}} \right)} \right]} \right\}:{\left[ {\left( {\dfrac{{ – 1}}{2}} \right) + 0,25} \right]^2}\)
\(\begin{array}{l} = \left\{ {\dfrac{{71}}{{18}}.\left[ {\dfrac{5}{2} – \dfrac{5}{9}} \right]} \right\}:{\left[ {\left( {\dfrac{{ – 1}}{2}} \right) + \dfrac{1}{4}} \right]^2}\\ = \left( {\dfrac{{71}}{{18}}.\dfrac{{35}}{{18}}} \right):{\left( {\dfrac{1}{4}} \right)^2} = \dfrac{{2485}}{{324}}:\dfrac{1}{{16}} \\= \dfrac{{2485}}{{324}}.16 = \dfrac{{9940}}{{81}}\end{array}\)