Bài 11 trang 12 SGK Toán 7 tập 1

Đề bài

Tính

a) \({{ - 2} \over 7}.{{21} \over 8}\)

b) \(0,24.{{ - 15} \over 4}\)

c) \(\left( { - 2} \right).\left( { - {7 \over {12}}} \right)\)

d) \(\left( { - {3 \over {25}}} \right):6.\)

Hướng dẫn giải

Áp dụng công thức nhân, chia hai số hữu tỉ:

\[\begin{array}{l}
\frac{a}{b}.\frac{c}{d} = \frac{{a.c}}{{b.d}}\\
\frac{a}{b}:\frac{c}{d} = \frac{{a.d}}{{b.c}}\;\;\left( {b,\;c,\;d \ne 0} \right).
\end{array}\]

Lời giải chi tiết

\(a)\;\frac{{ - 2}}{7}.\frac{{21}}{8} = \frac{{ - 2.21}}{{7.8}} ={{ - 2.3.7} \over {7.2.4}}\)\(= \frac{{ - 1.3}}{{1.4}} = - \frac{3}{4}.\)

\(b)\;0,24.\frac{{ - 15}}{4} = \frac{{24}}{{100}}.\frac{{ - 15}}{4} \)\(={{6.4} \over {25.4}}.{{ - 15} \over 4}\)\(= \frac{6}{{25}}.\frac{{ - 15}}{4}\)\(= \frac{{6.\left( { - 15} \right)}}{{25.4}}\)\(={{3.2.5.( - 3)} \over {5.5.2.2}} \)\( = \frac{{3.\left( { - 3} \right)}}{{5.2}} = \frac{{ - 9}}{{10}}.\)

\(c)\;\left( { - 2} \right).\left( { - \frac{7}{{12}}} \right) = \frac{{ - 2.\left( { - 7} \right)}}{{12}} \)\(={{2.7} \over {2.6}}\)\(= \frac{{1.7}}{6} = \frac{7}{6} = 1\frac{1}{6}.\)

\(d)\;\left( { - \frac{3}{{25}}} \right):6\; = \frac{{ - 3}}{{25}}.\frac{1}{6} = \frac{{ - 3}}{{25.6}} \)\(={{ - 3} \over {25.2.3}}\)\(= \frac{{ - 1}}{{25.2}} = - \frac{1}{{50}}\)