Giải bài 2 trang 15 SGK Toán 7 tập 1 – Chân trời sáng tạo

Đề bài

Tính:

a)\(0,75 – \frac{5}{6} + 1\frac{1}{2};\)                                    

b)\(\frac{3}{7} + \frac{4}{{15}} + \left( {\frac{{ – 8}}{{21}}} \right) + \left( { – 0,4} \right);\)

c)\(0,625 + \left( {\frac{{ – 2}}{7}} \right) + \frac{3}{8} + \left( {\frac{{ – 5}}{7}} \right) + 1\frac{2}{3}\)           

d)\(\left( { – 3} \right).\left( {\frac{{ – 38}}{{21}}} \right).\left( {\frac{{ – 7}}{6}} \right).\left( { – \frac{3}{{19}}} \right);\)

e) \(\left( {\frac{{11}}{{18}}:\frac{{22}}{9}} \right).\frac{8}{5};\)                                    

g)\(\left[ {\left( {\frac{{ – 4}}{5}} \right).\frac{5}{8}} \right]:\left( {\frac{{ – 25}}{{12}}} \right)\)

Lời giải chi tiết

a)

\(\begin{array}{l}0,75 – \frac{5}{6} + 1\frac{1}{2} = \frac{3}{4} – \frac{5}{6} + \frac{3}{2}\\ = \frac{9}{{12}} – \frac{{10}}{{12}} + \frac{{18}}{{12}} = \frac{{17}}{{12}}\end{array}\)                                    

b)

\(\begin{array}{l}\frac{3}{7} + \frac{4}{{15}} + \left( {\frac{{ – 8}}{{21}}} \right) + \left( { – 0,4} \right) = \frac{3}{7} + \frac{4}{{15}} – \frac{8}{{21}} – \frac{2}{5}\\ = \left( {\frac{3}{7} – \frac{8}{{21}}} \right) + \left( {\frac{4}{{15}} – \frac{2}{5}} \right)\\ = \left( {\frac{9}{{21}} – \frac{8}{{21}}} \right) + \left( {\frac{4}{{15}} – \frac{6}{{15}}} \right)\\ = \frac{1}{{21}} + \left( {\frac{{ – 2}}{{15}}} \right)\\ = \frac{5}{{105}} – \frac{{14}}{{105}}\\ = \frac{{ – 9}}{{105}} = \frac{{ – 3}}{{35}}\end{array}\)

c)

\(\begin{array}{l}0,625 + \left( {\frac{{ – 2}}{7}} \right) + \frac{3}{8} + \left( {\frac{{ – 5}}{7}} \right) + 1\frac{2}{3}\\ = \frac{5}{8} + \left( {\frac{{ – 2}}{7}} \right) + \frac{3}{8} – \frac{5}{7} + \frac{5}{3}\\ = \left( {\frac{5}{8} + \frac{3}{8}} \right) + \left( {\frac{{ – 2}}{7} – \frac{5}{7}} \right) + \frac{5}{3}\\ = 1 – 1 + \frac{5}{3} = \frac{5}{3}\end{array}\)          

 d)

\(\begin{array}{l}\left( { – 3} \right).\left( {\frac{{ – 38}}{{21}}} \right).\left( {\frac{{ – 7}}{6}} \right).\left( { – \frac{3}{{19}}} \right)\\ = \frac{{ – 3.\left( { – 38} \right).\left( { – 7} \right).\left( { – 3} \right)}}{{21.6.19}}\\ = \frac{{3.38.7.3}}{{21.6.19}}\\ = \frac{{3.2.19.7.3}}{{3.7.3.2.19}}\\ = 1\end{array}\)

e)

 \(\begin{array}{l}\left( {\frac{{11}}{{18}}:\frac{{22}}{9}} \right).\frac{8}{5} = \left( {\frac{{11}}{{18}}.\frac{9}{{22}}} \right).\frac{8}{5}\\ = \frac{{11.9.4.2}}{{9.2.2.11.5}} = \frac{2}{5}\end{array}\)                                   

 g)

\(\left[ {\left( {\frac{{ – 4}}{5}} \right).\frac{5}{8}} \right]:\left( {\frac{{ – 25}}{{12}}} \right) = \frac{{ – 20}}{{40}}:\left( {\frac{{ – 25}}{{12}}} \right)\\ = \frac{{ – 1}}{2}.\frac{{ – 12}}{{25}} = \frac{6}{{25}}\)