Tính:
\(a)\frac{2}{{15}} + \left( {\frac{{ – 5}}{{24}}} \right)\)
b) \(\left( {\frac{{ – 5}}{9}} \right) – \left( { – \frac{7}{{27}}} \right);\)
c)\(\left( { – \frac{7}{{12}}} \right) + 0,75\)
d)\(\left( {\frac{{ – 5}}{9}} \right) – 1,25\)
e)\(0,34.\frac{{ – 5}}{{17}}\)
g) \(\frac{4}{9}:\left( { – \frac{8}{{15}}} \right);\)
h)\(\left( {1\frac{2}{3}} \right):\left( {2\frac{1}{2}} \right)\)
i) \(\frac{2}{5}.\left( { – 1.25} \right)\)
k) \(\left( {\frac{{ – 3}}{5}} \right).\left( {\frac{{15}}{{ – 7}}} \right).3\frac{1}{9}\)
Phương pháp giải
– Đưa các số về phân số
– Thực hiện quy tắc cộng, trừ, nhân, chia số hữu tỉ.
Lời giải chi tiết
\(a)\frac{2}{{15}} + \left( {\frac{{ – 5}}{{24}}} \right) = \frac{{16}}{{120}} + \left( {\frac{{ – 25}}{{120}}} \right) = \frac{{ – 9}}{{120}} = \frac{{ – 3}}{{40}}\)
b) \(\left( {\frac{{ – 5}}{9}} \right) – \left( { – \frac{7}{{27}}} \right) = \left( {\frac{{ – 15}}{{27}}} \right) + \frac{7}{{27}} = \frac{{ – 8}}{{27}}\)
c)\(\left( { – \frac{7}{{12}}} \right) + 0,75 = \left( { – \frac{7}{{12}}} \right) + \frac{3}{4} = \left( { – \frac{7}{{12}}} \right) + \frac{9}{{12}} = \frac{2}{{12}} = \frac{1}{6}\)
d)\(\left( {\frac{{ – 5}}{9}} \right) – 1,25 = \left( {\frac{{ – 5}}{9}} \right) – \frac{5}{4} = \left( {\frac{{ – 20}}{{36}}} \right) – \frac{{45}}{{36}} = \frac{{ – 65}}{{36}}\)
e)\(0,34.\frac{{ – 5}}{{17}} = \frac{{17}}{{50}}.\frac{{ – 5}}{{17}} = \frac{{ – 1}}{{10}}\)
g) \(\frac{4}{9}:\left( { – \frac{8}{{15}}} \right) = \frac{4}{9}.\left( { – \frac{{15}}{8}} \right) = \frac{{ – 5}}{6}\)
h)\(\left( {1\frac{2}{3}} \right):\left( {2\frac{1}{2}} \right) = \frac{5}{3}:\frac{5}{2} = \frac{5}{3}.\frac{2}{5} = \frac{2}{3}\)
i) \(\frac{2}{5}.\left( { – 1.25} \right) = \frac{2}{5}.\frac{{ – 5}}{4} = \frac{{ – 1}}{2}\)
k) \(\left( {\frac{{ – 3}}{5}} \right).\left( {\frac{{15}}{{ – 7}}} \right).3\frac{1}{9} = \left( {\frac{{ – 3}}{5}} \right).\left( {\frac{{15}}{{ – 7}}} \right).\frac{{28}}{9} = \frac{{ – 3.3.5.7.4}}{{5.\left( { – 7} \right).3.3}} = 4\)
Để lại một bình luận