Tính:
a) \(\dfrac{3}{{10}} + \left( {\dfrac{{ - 5}}{{12}}} \right)\)
b) \(\dfrac{{ - 3}}{8} - \left( { - \dfrac{7}{{24}}} \right)\)
c) \(\left( {\dfrac{{ - 5}}{{14}}} \right) + 0,25\)
d) \(\left( { - \dfrac{7}{8}} \right) - 1,25\)
e) \(\left( {\dfrac{{ - 5}}{{14}}} \right).\dfrac{{21}}{{25}}\)
Advertisements (Quảng cáo)
f) \(\dfrac{8}{{27}}:\left( { - \dfrac{{16}}{{45}}} \right)\)
g) \(\left( {1\dfrac{5}{6}} \right):\left( {4\dfrac{1}{8}} \right)\)
h) \(0,38.\left( { - \dfrac{7}{{19}}} \right)\)
i) \(\left( {\dfrac{{ - 4}}{5}} \right).\left( {\dfrac{{15}}{{ - 8}}} \right).1\dfrac{1}{9}\)
Ta có thể viết các số hữu tỉ dưới dạng phân số rồi sau đó áp dụng qui tắc cộng, trừ phân số .
\(\begin{array}{l}a)\dfrac{3}{{10}} + \left( {\dfrac{{ - 5}}{{12}}} \right) = \dfrac{{18}}{{60}} + \left( {\dfrac{{ - 25}}{{60}}} \right) = \dfrac{{ - 7}}{{60}}\\b)\dfrac{{ - 3}}{8} - \left( { - \dfrac{7}{{24}}} \right) = \dfrac{{ - 3}}{8} + \dfrac{7}{{24}} = \dfrac{{ - 9}}{{24}} + \dfrac{7}{{24}} = \dfrac{{ - 2}}{{24}} = \dfrac{{ - 1}}{{12}}\\c)\left( {\dfrac{{ - 5}}{{14}}} \right) + 0,25 = \left( {\dfrac{{ - 5}}{{14}}} \right) + \dfrac{1}{4} = \left( {\dfrac{{ - 10}}{{28}}} \right) + \dfrac{7}{{28}} = \dfrac{{ - 3}}{{28}}\\d)\left( { - \dfrac{7}{8}} \right) - 1,25 = - \left( {\dfrac{7}{8} + \dfrac{5}{4}} \right) = - \left( {\dfrac{7}{8} + \dfrac{{10}}{8}} \right) = \dfrac{{ - 17}}{8}\\e)\left( {\dfrac{{ - 5}}{{14}}} \right).\dfrac{{21}}{{25}} = - \dfrac{{5.21}}{{14.25}} = - \dfrac{{105}}{{350}} = - \dfrac{{105:35}}{{350:35}} = - \dfrac{3}{{10}}\\f)\dfrac{8}{{27}}:\left( { - \dfrac{{16}}{{45}}} \right) = \dfrac{8}{{27}}.\left( { - \dfrac{{45}}{{16}}} \right) = - \dfrac{{8.45}}{{27.16}} = - \dfrac{{8.9.5}}{{9.3.8.2}} = - \dfrac{5}{6}\\g)\left( {1\dfrac{5}{6}} \right):\left( {4\dfrac{1}{8}} \right) = \dfrac{{11}}{6}:\dfrac{{33}}{8} = \dfrac{{11}}{6}.\dfrac{8}{{33}} = \dfrac{{11.2.4}}{{2.3.11.3}} = \dfrac{4}{9}\\h)0,38.\left( { - \dfrac{7}{{19}}} \right) = \dfrac{{38}}{{100}}.\left( { - \dfrac{7}{{19}}} \right) = - \dfrac{{2.19.7}}{{2.50.19}} = \dfrac{{ - 7}}{{50}}\\i)\left( {\dfrac{{ - 4}}{5}} \right).\left( {\dfrac{{15}}{{ - 8}}} \right).1\dfrac{1}{9} = \dfrac{4}{5}.\dfrac{{15}}{8}.\dfrac{{10}}{9} = \dfrac{{4.15.10}}{{5.8.9}} = \dfrac{{4.3.5.5.2}}{{5.4.2.3.3}} = \dfrac{5}{2}\end{array}\)