Bài 51 trang 53 Vở bài tập toán 6 tập 2

Đề bài

Tính:

a) \( \displaystyle 5.{{ - 3} \over {10}}\)

b) \( \displaystyle {2 \over 7} + {5 \over 7}.{{14} \over {25}}\)

c) \( \displaystyle {1 \over 3} - {5 \over 4}.{4 \over {15}}\)

d) \( \displaystyle \left( {{3 \over 4} + {{ - 7} \over 2}} \right).\left( {{2 \over {11}} + {{12} \over {22}}} \right)\) 

Lời giải chi tiết

\(\begin{array}{l}
a)\,\,5.\dfrac{{ - 3}}{{10}} = \dfrac{{5.( - 3)}}{{10}}=\dfrac{{1. (- 3)}}{{2}}\\=\dfrac{{ - 3}}{2}\\
b)\,\,\dfrac{2}{7} + \dfrac{5}{7}.\dfrac{{14}}{{25}} = \dfrac{2}{7} + \dfrac{{5.14}}{{7.25}} \\= \dfrac{2}{7} + \dfrac{{1.2}}{{1.5}} = \dfrac{{2}}{{7}} + \dfrac{{2}}{{5}} = \dfrac{{24}}{{35}}\\
c)\,\dfrac{1}{3} - \dfrac{5}{4}.\dfrac{4}{{15}} = \dfrac{1}{3} - \dfrac{5.4}{4.15}\\=\dfrac{1}{3} -\dfrac{1.1}{1.3} = 0\\
d)\,\,\left( {\dfrac{3}{4} + \dfrac{{ - 7}}{2}} \right).\left( {\dfrac{2}{{11}} + \dfrac{{12}}{{22}}} \right)\\ = \left( {\dfrac{3}{4} + \dfrac{{ - 14}}{4}} \right).\left( {\dfrac{4}{{22}} + \dfrac{{12}}{{22}}} \right) \\= \dfrac{{ - 11}}{4}.\dfrac{{16}}{{22}} = \dfrac{{ (- 11).16}}{4.22} =\dfrac{{(-1).4}}{{1.2}}\\ =\dfrac{-4}{2}=- 2 
\end{array}\)