Tính:
a)\(\frac{3}{7}.\left( { – \frac{1}{9}} \right) + \frac{3}{7}.\left( { – \frac{2}{3}} \right);\)
b)\(\left( {\frac{{ – 7}}{{13}}} \right).\frac{5}{{12}} + \left( {\frac{{ – 7}}{{13}}} \right).\frac{7}{{12}} + \left( {\frac{{ – 6}}{{13}}} \right);\)
c)\(\left[ {\left( {\frac{{ – 2}}{3} + \frac{3}{7}} \right)} \right]:\frac{5}{9} + \left( {\frac{4}{7} – \frac{1}{3}} \right):\frac{5}{9}\)
d)\(\frac{5}{9}:\left( {\frac{1}{{11}} – \frac{5}{{22}}} \right) + \frac{5}{9}:\left( {\frac{1}{{15}} – \frac{2}{3}} \right);\)
e) \(\frac{3}{5} + \frac{3}{{11}} – \left( {\frac{{ – 3}}{7}} \right) + \left( {\frac{{ – 2}}{{97}}} \right) – \frac{1}{{35}} – \frac{3}{4} + \left( {\frac{{ – 23}}{{44}}} \right)\)
Phương pháp giải
– Áp dụng tính chất phân phối của phép nhân với phép cộng: a.c+b.c=(a+b).c
– Áp dụng tính chất giao hoán của phép cộng
Lời giải chi tiết
a)
\(\begin{array}{l}\frac{3}{7}.\left( { – \frac{1}{9}} \right) + \frac{3}{7}.\left( { – \frac{2}{3}} \right)\\ = \frac{3}{7}.\left( { – \frac{1}{9} + \frac{-2}{3}} \right)\\ = \frac{3}{7}.\left( { – \frac{1}{9} – \frac{6}{9}} \right)\\ = \frac{3}{7}.\frac{{ – 7}}{9} = \frac{{ – 1}}{3}\end{array}\)
b)
\(\begin{array}{l}\left( {\frac{{ – 7}}{{13}}} \right).\frac{5}{{12}} + \left( {\frac{{ – 7}}{{13}}} \right).\frac{7}{{12}} + \left( {\frac{{ – 6}}{{13}}} \right)\\ = \frac{{ – 7}}{{13}}.\left( {\frac{5}{{12}} + \frac{7}{{12}}} \right) + \left( {\frac{{ – 6}}{{13}}} \right)\\ = \frac{{ – 7}}{{13}}.1 + \left( {\frac{{ – 6}}{{13}}} \right)\\ = \frac{{ – 7}}{{13}} + \left( {\frac{{ – 6}}{{13}}} \right)\\ = \frac{{ – 13}}{{13}}\\ = -1\end{array}\)
c)
\(\begin{array}{l}\left[ {\left( {\frac{{ – 2}}{3} + \frac{3}{7}} \right)} \right]:\frac{5}{9} + \left( {\frac{4}{7} – \frac{1}{3}} \right):\frac{5}{9}\\ = \left[ {\left( {\frac{{ – 2}}{3} + \frac{3}{7}} \right)} \right].\frac{9}{5} + \left( {\frac{4}{7} – \frac{1}{3}} \right).\frac{9}{5}\\ = \left( {\frac{{ – 2}}{3} + \frac{3}{7} + \frac{4}{7} – \frac{1}{3}} \right).\frac{9}{5}\\ = \left[ {\left( {\frac{{ – 2}}{3} – \frac{1}{3}} \right) + \left( {\frac{3}{7} + \frac{4}{7}} \right)} \right].\frac{9}{5}\\ = \left( { – 1 + 1} \right).\frac{9}{5}\\ = 0.\frac{9}{5} = 0\end{array}\)
d)
\(\begin{array}{l}\frac{5}{9}:\left( {\frac{1}{{11}} – \frac{5}{{22}}} \right) + \frac{5}{9}:\left( {\frac{1}{{15}} – \frac{2}{3}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} – \frac{5}{{22}}} \right) + \frac{5}{9}:\left( {\frac{1}{{15}} – \frac{{10}}{{15}}} \right)\\ = \frac{5}{9}:\frac{{ – 3}}{{22}} + \frac{5}{9}:\frac{{ – 9}}{15}\\= \frac{5}{9}:\frac{{ – 3}}{{22}} + \frac{5}{9}:\frac{{ – 3}}{5}\\ = \frac{5}{9}.\frac{{ – 22}}{3} + \frac{5}{9}.\frac{{ – 5}}{3}\\ = \frac{5}{9}.\left( {\frac{{ – 22}}{3} – \frac{5}{3}} \right)\\ = \frac{5}{9}.\frac{-27}{3}= \frac{5}{9}.\left( { – 9} \right) = – 5\end{array}\)
e)
\(\begin{array}{l}\frac{3}{5} + \frac{3}{{11}} – \left( {\frac{{ – 3}}{7}} \right) + \left( {\frac{{ – 2}}{{97}}} \right) – \frac{1}{{35}} – \frac{3}{4} + \left( {\frac{{ – 23}}{{44}}} \right)\\ = \frac{3}{5} + \frac{3}{{11}} + \frac{3}{7} – \frac{2}{{97}} – \frac{1}{{35}} – \frac{3}{4} – \frac{{23}}{{44}}\\ = \left( {\frac{3}{5} + \frac{3}{7} – \frac{1}{{35}}} \right) + \left( {\frac{3}{{11}} – \frac{3}{4} – \frac{{23}}{{44}}} \right) – \frac{2}{{97}}\\ = \left( {\frac{{21}}{{35}} + \frac{{15}}{{35}} – \frac{1}{{35}}} \right) + \left( {\frac{{12}}{{44}} – \frac{{33}}{{44}} – \frac{{23}}{{44}}} \right) – \frac{2}{{97}}\\ = 1 + \left( { – 1} \right) – \frac{2}{{97}}\\ = – \frac{2}{{97}}\end{array}\)
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