Bài 8 trang 10 SGK Toán 7 tập 1

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

Lời giải

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


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