Giải bài 1 trang 15 SGK Toán 7 tập 1 - Chân trời sáng tạo

Đề bài

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) \(\left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).3\frac{1}{9}\)

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) \(\left( { - \frac{7}{{12}}} \right) + 0,75 \) \(= \left( { - \frac{7}{{12}}} \right) + \frac{75}{100} \) \(= \left( { - \frac{7}{{12}}} \right) + \frac{3}{4} \) \(= \left( { - \frac{7}{{12}}} \right) + \frac{9}{{12}} \) \(= \frac{2}{{12}} = \frac{1}{6}\)

d) \(\left( {\frac{{ - 5}}{9}} \right) - 1,25 \) \(=\left( {\frac{{ - 5}}{9}} \right) - \frac{125}{100} = \left( {\frac{{ - 5}}{9}} \right) - \frac{5}{4}\) \( = \left( {\frac{{ - 20}}{{36}}} \right) - \frac{{45}}{{36}} = \frac{{ - 65}}{{36}}\)

e) \(0,34.\frac{{ - 5}}{{17}} =\frac{{34}}{{100}}.\frac{{ - 5}}{{17}}\) \(= \frac{{17}}{{50}}.\frac{{ - 5}}{{17}}\) \(= \frac{{ - 1}}{{10}}\)

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) \(\frac{2}{5}.\left( { - 1,25} \right)\) \(= \frac{2}{5}.\frac{{ - 125}}{100}\) \(= \frac{2}{5}.\frac{{ - 5}}{4}\) \(= \frac{{ - 1}}{2}\)

k) \(\left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).3\frac{1}{9} \) \(= \left( {\frac{{ - 3}}{5}} \right).\left( {\frac{{15}}{{ - 7}}} \right).\frac{{28}}{9}\) \( = \frac{{ - 3.3.5.7.4}}{{5.\left( { - 7} \right).3.3}} = 4\)