Giải bài 16 trang 13 - Sách giáo khoa Toán 7 tập 1

Đề bài

Tính

Hướng dẫn giải

a)  $(\dfrac{-2}{3} + \dfrac{3}{7}): \dfrac{4}{5} + (\dfrac{-1}{3} + \dfrac{4}{7}) : \dfrac{4}{5} = (\dfrac{-2}{3}+ \dfrac{3}{7}+ \dfrac{-1}{3} +\dfrac{4}{7}):\dfrac{4}{5}$

  $= (\dfrac{-3}{3} +\dfrac{7}{7}): \dfrac{4}{5} = (-1+1):\dfrac{4}{5} = 0$

b)  $\dfrac{5}{9}: (\dfrac{1}{11} – \dfrac{5}{22}) + \dfrac{5}{9} :(\dfrac{1}{15} – \dfrac{2}{3}) = \dfrac{5}{9}: \dfrac{2 – 5}{22} + \dfrac{5}{9}: \dfrac{1-10}{15}$ 

  $= \dfrac{5}{9}.\dfrac{22}{-3}+\dfrac{5}{9}.\dfrac{15}{-9} = \dfrac{5}{9}(\dfrac{22}{-3} + \dfrac{15}{-9})=\dfrac{5}{9}.\dfrac{-27}{3}= 5.(-1)=-5$