Giải bài 4 trang 15 SGK Toán 7 tập 1 – Chân trời sáng tạo

Đề bài

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) \(\frac{3}{5} + \frac{3}{{11}} – \left( {\frac{{ – 3}}{7}} \right) + \left( {\frac{{ – 2}}{{97}}} \right) – \frac{1}{{35}} – \frac{3}{4} + \left( {\frac{{ – 23}}{{44}}} \right)\)

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)

\(\begin{array}{l}\frac{3}{5} + \frac{3}{{11}} – \left( {\frac{{ – 3}}{7}} \right) + \left( {\frac{{ – 2}}{{97}}} \right) – \frac{1}{{35}} – \frac{3}{4} + \left( {\frac{{ – 23}}{{44}}} \right)\\ = \frac{3}{5} + \frac{3}{{11}} + \frac{3}{7} – \frac{2}{{97}} – \frac{1}{{35}} – \frac{3}{4} – \frac{{23}}{{44}}\\ = \left( {\frac{3}{5} + \frac{3}{7} – \frac{1}{{35}}} \right) + \left( {\frac{3}{{11}} – \frac{3}{4} – \frac{{23}}{{44}}} \right) – \frac{2}{{97}}\\ = \left( {\frac{{21}}{{35}} + \frac{{15}}{{35}} – \frac{1}{{35}}} \right) + \left( {\frac{{12}}{{44}} – \frac{{33}}{{44}} – \frac{{23}}{{44}}} \right) – \frac{2}{{97}}\\ = \frac{35}{{35}}+ \frac{-44}{{44}}- \frac{2}{{97}}\\= 1 + \left( { – 1} \right) – \frac{2}{{97}}\\ =  – \frac{2}{{97}}\end{array}\)