Bài 11 trang 12 SGK Toán 7 tập 1

Đề bài

Tính

\(\eqalign{
& a)\,\,{{ - 2} \over 7}.{{21} \over 8} \cr
& b)\,\,0,24.{{ - 15} \over 4} \cr
& c)\,\,\left( { - 2} \right).\left( { - {7 \over {12}}} \right) \cr
& d)\,\,\left( { - {3 \over {25}}} \right):6 \cr} \)

Lời giải

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

\(b)\;0,24.\dfrac{{ - 15}}{4} = \dfrac{{24}}{{100}}.\dfrac{{ - 15}}{4} \)\( = \dfrac{{6.4}}{{25.4}}.\dfrac{{ - 15}}{4}\)\(= \dfrac{6}{{25}}.\dfrac{{ - 15}}{4}= \dfrac{{6.\left( { - 15} \right)}}{{25.4}}\)

\(= \dfrac{{3.2.5.( - 3)}}{{5.5.2.2}} \)\( = \dfrac{{3.\left( { - 3} \right)}}{{5.2}} = \dfrac{{ - 9}}{{10}}.\)

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

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


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