Bài 4 trang 9 Vở bài tập toán 7 tập 1

Đề bài

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 chi tiết

\(\begin{array}{l}
a)\;\dfrac{3}{7} + \left( { - \dfrac{5}{2}} \right) + \left( { - \dfrac{3}{5}} \right) \\ = \dfrac{3}{7} + \dfrac{{ - 5}}{2} + \dfrac{{ - 3}}{5} \\= \dfrac{{3.2.5}}{{70}} + \dfrac{{ - 5.7.5}}{{70}} + \dfrac{{ - 3.7.2}}{{70}}\\= \dfrac{{30}}{{70}} + \dfrac{{ - 175}}{{70}} + \dfrac{{ - 42}}{{70}}\\=  \dfrac{{-187}}{{70}} = - 2\dfrac{{47}}{{70}}.\\
b)\;\;\left( { - \dfrac{4}{3}} \right) + \left( { - \dfrac{2}{5}} \right) + \left( { - \dfrac{3}{2}} \right) \\  = \dfrac{{ - 4}}{3} + \dfrac{{ - 2}}{5} + \dfrac{{ - 3}}{2}\\ = \dfrac{{ - 4.5.2}}{{30}} + \dfrac{{ - 2.3.2}}{{30}} + \dfrac{{ - 3.3.5}}{{30}}\\= \dfrac{{ - 40}}{{30}} + \dfrac{{ - 12}}{{30}} + \dfrac{{ - 45}}{{30}}\\ = - \dfrac{{97}}{{30}} = - 3\dfrac{7}{{30}}.\\
c)\;\dfrac{4}{5} - \left( { - \dfrac{2}{7}} \right) - \dfrac{7}{{10}}\\ = \dfrac{4}{5} - \dfrac{{ - 2}}{7} - \dfrac{7}{{10}} \\= \dfrac{{4.14}}{{70}} - \dfrac{{ - 2.10}}{{70}} - \dfrac{{7.7}}{{70}}\\= \dfrac{{56}}{{70}} - \dfrac{{-20}}{{70}} - \dfrac{{49}}{{70}}\\
= \dfrac{{56 -(- 20) - 49}}{{70}} = \dfrac{{27}}{{70}}.
\end{array}\)

\(\eqalign{
& d)\;{2 \over 3} - \left[ {\left( { - {7 \over 4}} \right) - \left( {{1 \over 2} + {3 \over 8}} \right)} \right] \cr
& = {2 \over 3} - \left[ { - {7 \over 4} - \left( {{4 \over 8} + {3 \over 8}} \right)} \right] \cr
& = {2 \over 3} - \left[ { - {7 \over 4} - {7 \over 8}} \right] = {2 \over 3} - \left( {{{ - 14} \over 8} - {7 \over 8}} \right) \cr
& = {2 \over 3} - {{ - 21} \over 8} = {{16} \over {24}} - {{ - 63} \over {24}} = {{79} \over {24}} = 3{7 \over {24}} \cr} \)