Bài 114 trang 50 SGK Toán 6 tập 2

Đề bài

Tính:

\(\displaystyle \left( { - 3,2} \right).{{ - 15} \over {64}} + \left( {0,8 - 2{4 \over {15}}} \right):3{2 \over 3}.\)

Lời giải chi tiết

\( \displaystyle \left( { - 3,2} \right).{{ - 15} \over {64}} + \left( {0,8 - 2{4 \over {15}}} \right):3{2 \over 3}.\)

\( \displaystyle = {{ - 32} \over {10}}.{{ - 15} \over {64}} + \left( {{8 \over {10}} - {{34} \over {15}}} \right):{{11} \over 3}\)

\(\begin{array}{l}
= \dfrac{{ - 32.\left( { - 15} \right)}}{{10.64}} + \left( {\dfrac{4}{5} - \dfrac{{34}}{{15}}} \right):\dfrac{{11}}{3}\\
= \dfrac{{\left( { - 1} \right).\left( { - 3} \right)}}{{2.2}} + \left( {\dfrac{{12}}{{15}} - \dfrac{{34}}{{15}}} \right):\dfrac{{11}}{3}\\
= \dfrac{3}{4} + \dfrac{{ - 22}}{{15}}:\dfrac{{11}}{3}\\
= \dfrac{3}{4} + \dfrac{{ - 22}}{{15}}.\dfrac{3}{{11}}\\
= \dfrac{3}{4} + \dfrac{{\left( { - 22} \right).3}}{{15.11}}\\
= \dfrac{3}{4} + \dfrac{{\left( { - 2} \right).1}}{{5.1}}\\
= \dfrac{3}{4} + \dfrac{{ - 2}}{5}\\
= \dfrac{{15}}{{20}} + \dfrac{{ - 8}}{{20}} = \dfrac{{15 + \left( { - 8} \right)}}{{20}}\\
= \dfrac{7}{{20}}
\end{array}\)