Giải bài 1 trang 88 - SGK Toán lớp 7 Tập 2

a) \(9,6.2\dfrac{1}{2} - \left(2.125 - 1\dfrac{5}{12} \right) : \dfrac{1}{4}\)
\(= 9,6.\dfrac{5}{2} - \left(250 - \dfrac{17}{12} \right) : \dfrac{1}{4}\)
\(= \dfrac{48}{2} - \left(\dfrac{3000}{12} - \dfrac{17}{12} \right) : \dfrac{1}{4}\)
\(= 24 - \dfrac{2983}{12} . 4\)
\(= 24 - \dfrac{2983}{3}\)
\(= \dfrac{72}{3} - \dfrac{2983}{3}\)
\(= \dfrac{-2911}{3} = -970\dfrac{1}{3}\)
b) \(\dfrac{5}{18} - 1,456 : \dfrac{7}{25} + 4,5.\dfrac{4}{5}\)
\(= \dfrac{5}{18} - 1,456 . \dfrac{25}{7} + 4,5.\dfrac{4}{5}\)
\(= \dfrac{5}{18} - \dfrac{36,4}{7} + \dfrac{18}{5}\)
\(= 0,28 - 5,2 + 3,6\)
\(= -1,32\)
c) \(\left(\dfrac{1}{2} + 0,8 - 1\dfrac{1}{3} \right).\left(2,3 + 4\dfrac{7}{25} - 1,28 \right)\)
\(= \left(\dfrac{1}{2} + \dfrac{4}{5} - \dfrac{4}{3} \right).\left(\dfrac{23}{10} + \dfrac{107}{25} - \dfrac{32}{25} \right)\)
\(= \left(\dfrac{15}{30} + \dfrac{24}{30} - \dfrac{40}{30} \right).\left(\dfrac{115}{50} + \dfrac{214}{50} - \dfrac{64}{50} \right)\)
\(= \dfrac{-1}{30} - \dfrac{265}{50}\)

\(= \dfrac{-1}{30} - \dfrac{53}{10}\)

\(= \dfrac{-1}{30} - \dfrac{159}{30}\)

\( = \dfrac{-160}{30} = \dfrac{-16}{3}\)

d) \((-5).12:\left[\left( -\dfrac{1}{4}\right) + \dfrac{1}{2} : (-2)\right] + 1\dfrac{1}{3}\)
\(= -60 : \left[ -\dfrac{1}{4}- \dfrac{1}{4}\right] + \dfrac{4}{3}\)
\(= -60 : \left( -\dfrac{1}{2}\right) + \dfrac{4}{3}\)
\(= -60.(-2) + \dfrac{4}{3}\)
\(= 120 + \dfrac{4}{3}\)
\(= \dfrac{360}{3} + \dfrac{4}{3}\)
\(= \dfrac{364}{3}\)
\(= 121\dfrac{1}{3}\)

Lưu ý: Cách đưa hỗn số về dạng phân số: \(a\dfrac{b}{c} = \dfrac{a.c + b}{c}\)