Bài 8 trang 10 SGK Toán 7 tập 1

Đề bài

Tính:

\(\begin{array}{l}
a)\;\frac{3}{7} + \left( { - \frac{5}{2}} \right) + \left( { - \frac{3}{5}} \right)\\
b)\;\;\left( { - \frac{4}{3}} \right) + \left( { - \frac{2}{5}} \right) + \left( { - \frac{3}{2}} \right)\\
c)\;\frac{4}{5} - \left( { - \frac{2}{7}} \right) - \frac{7}{{10}}\\
d)\;\frac{2}{3} - \left[ {\left( { - \frac{7}{4}} \right) - \left( {\frac{1}{2} + \frac{3}{8}} \right)} \right].
\end{array}\)

Hướng dẫn giải

Áp dụng các công thức cộng, trừ hai số hữu tỉ: Với \(a,\;b,\;m \in Z,\;\;m > 0\) ta có:

\[\begin{array}{l}
\frac{a}{m} + \frac{b}{m} = \frac{{a + b}}{m}\\
\frac{a}{m} - \frac{b}{m} = \frac{{a - b}}{m}
\end{array}\]

Lời giải chi tiết

\(\begin{array}{l}
a)\;\frac{3}{7} + \left( { - \frac{5}{2}} \right) + \left( { - \frac{3}{5}} \right) \\= \frac{{30}}{{70}} + \frac{{ - 175}}{{70}} + \frac{{ - 42}}{{70}}\\
= \frac{{30 - 175 - 42}}{{70}} = - \frac{{187}}{{70}} = - 2\frac{{47}}{{40}}.\\
b)\;\;\left( { - \frac{4}{3}} \right) + \left( { - \frac{2}{5}} \right) + \left( { - \frac{3}{2}} \right) \\ = \frac{{ - 40}}{{30}} + \frac{{ - 12}}{{30}} + \frac{{ - 45}}{{30}}\\
= \frac{{ - 40 + \left( { - 12} \right) + \left( { - 45} \right)}}{{30}} = - \frac{{97}}{{30}} = - 3\frac{7}{{30}}.\\
c)\;\frac{4}{5} - \left( { - \frac{2}{7}} \right) - \frac{7}{{10}} = \frac{{56}}{{70}} + \frac{{20}}{{70}} - \frac{{49}}{{70}}\\
= \frac{{56 + 20 - 49}}{{70}} = \frac{{27}}{{70}}.\\
d)\;\frac{2}{3} - \left[ {\left( { - \frac{7}{4}} \right) - \left( {\frac{1}{2} + \frac{3}{8}} \right)} \right] \\ = \frac{2}{3} - \left[ {\left( { - \frac{7}{4}} \right) - \left( {\frac{4}{8} + \frac{3}{8}} \right)} \right]\\
= \frac{2}{3} - \left[ { - \frac{7}{4} - \frac{7}{8}} \right] = \frac{2}{3} - \left( { - \frac{{14}}{8} - \frac{7}{8}} \right)\\
= \frac{2}{3} - \left( { - \frac{{21}}{8}} \right) = \frac{2}{3}+\frac{21}{8}\\= \frac{{16}}{{24}} + \frac{{63}}{{24}} = \frac{{79}}{{24}}.
\end{array}\)