Giải bài 13 trang 12 - Sách giáo khoa Toán 7 tập 1

Đề bài

Tính

Hướng dẫn giải

\(a) \dfrac{-3}{4}.\dfrac{12}{-5}.(-\dfrac{25}{6})=-\dfrac{3}{4}.\dfrac{12}{-5}.\dfrac{-25}{6}=\dfrac{(-3).12.(-25)}{4.(-5).6}=\dfrac{(-3).2.5}{4.1.1}=\dfrac{-15}{2}=-7\dfrac{1}{2}\);

\(b) (-2)\dfrac{-38}{21}.\dfrac{-7}{4}.(-\dfrac{3}{8})=(-2).\dfrac{-38}{21}.\dfrac{-7}{4}.\dfrac{-3}{8} \)

 \(= \dfrac{(-2).(-28).(-7).(-3)}{21.4.8} = \dfrac{(-1).(-19).1}{1.2.4}= \dfrac{19}{8}=2\dfrac{3}{8} ;\)

\(c) (\dfrac{11}{12}:\dfrac{33}{16}).\dfrac{3}{5}=\dfrac{11}{12}.\dfrac{16}{33}.\dfrac{3}{5}=\dfrac{11.16.3}{12.33.5}=\dfrac{1.4.3}{3.3.5}=\dfrac{4}{15}\);

\(d) \dfrac{7}{23}.[(\dfrac{-8}{6})-\dfrac{45}{18}]=\dfrac{7}{23}.[\dfrac{-8}{6}-\dfrac{15}{6}]=\dfrac{7}{23}.\dfrac{-23}{6}=\dfrac{7.(-23)}{23.6}=\dfrac{-7}{6}=-1\dfrac{1}{6}.\)