Tính một cách hợp lí:
a) \(\dfrac{{ – 3}}{{11}} + 0,35 – \dfrac{8}{{11}}\);
b) \(\dfrac{9}{{19}} – 1,251 + \dfrac{{10}}{{19}} + 1,251\);
c) \( – 8.\dfrac{{24}}{7}.0,125\);
d) \(0,25.\dfrac{7}{{15}} – \dfrac{1}{4}.\dfrac{{ – 8}}{{15}} + 2,75\).
Tùy theo mỗi phép tính, ta áp dụng các tính chất giao hoán, kết hợp để tính một cách hợp lí.
a) \(\dfrac{{ – 3}}{{11}} + 0,35 – \dfrac{8}{{11}} = \left( {\dfrac{{ – 3}}{{11}} – \dfrac{8}{{11}}} \right) + 0,35 \\= \dfrac{{ – 11}}{{11}} + 0,35 = – 1 + 0,35 = – 0,65\)
b) \(\dfrac{9}{{19}} – 1,251 + \dfrac{{10}}{{19}} + 1,251\\ = \left( {\dfrac{9}{{19}} + \dfrac{{10}}{{19}}} \right) + \left( { – 1,251 + 1,251} \right)\\ =\dfrac{19}{19}+0= 1 + 0 = 1\)
c) \( – 8.\dfrac{{24}}{7}.0,125 = – 8.\dfrac{{24}}{7}.\dfrac{125}{1000} \\=- 8.\dfrac{{24}}{7}.\dfrac{1}{8} \\= \left( { – 8.\dfrac{1}{8}} \right).\dfrac{{24}}{7} = (- 1).\dfrac{{24}}{7} = \dfrac{{ – 24}}{7}\)
d)
\(\begin{array}{l}0,25.\dfrac{7}{{15}} – \dfrac{1}{4}.\dfrac{{ – 8}}{{15}} + 2,75\\ = \dfrac{25}{100}.\dfrac{7}{{15}} + \dfrac{1}{4}.\dfrac{8}{{15}} + \dfrac{{275}}{100}\\ =\dfrac{1}{4}.\dfrac{7}{{15}} + \dfrac{1}{4}.\dfrac{8}{{15}} + \dfrac{{275}}{100}\\ = \dfrac{1}{4}.\left( {\dfrac{7}{{15}} + \dfrac{8}{{15}}} \right) + \dfrac{{11}}{4}\\ = \dfrac{1}{4}.\dfrac{15}{15} + \dfrac{{11}}{4}\\= \dfrac{1}{4}.1 + \dfrac{{11}}{4}\\=\dfrac{1}{4} + \dfrac{{11}}{4} = \dfrac{{12}}{4} = 3\end{array}\).