Bài 2 (Trang 15, SGK Toán 7, Tập 1, Bộ Chân trời sáng tạo)

Tính:

$$\begin{gathered} a) 0,75 - \frac{5}{6} + 1\frac{1}{2}; \\ b) \frac{3}{7} + \frac{4}{15} + \left(\frac{-8}{21}\right) + (-0,4); \\ c) 0,625 + (\frac{-2}{7}) + \frac{3}{8} + (\frac{-5}{7}) + 1\frac{2}{3}; \\ d) (-3) . \left(\frac{-38}{21}\right).\left(\frac{-7}{6}\right).(-\frac{3}{19}); \\ e) (\frac{11}{18}:\frac{22}{9}).\frac{8}{5}; \end{gathered}$$

$g) [\left(\frac{-4}{5}\right).\frac{5}{8}]:(\frac{-25}{12}).$

Hướng dẫn giải:

$$\begin{gathered} a) 0,75 - \frac{5}{6} + 1\frac{1}{2} = \frac{75}{100} - \frac{5}{6} + \frac{3}{2} = \frac{3}{4} - \frac{5}{6} + \frac{3}{2} \\ b) \frac{3}{7} + \frac{4}{15} + \left(\frac{-8}{21}\right) + (-0,4) = \frac{3}{7} + \left(\frac{-8}{21}\right) + \frac{4}{15} + \frac{-4}{10} \\ =(\frac{9}{21} + \frac{-8}{21}) + \frac{4}{15} + \frac{-2}{5} \\ =\frac{1}{21} + \frac{4}{15} + \frac{-6}{15} = \frac{1}{21} + \frac{-2}{15} \\ =\frac{5}{105} + \frac{-14}{105} = \frac{-9}{105} = \frac{-3}{35} \\ c) 0,625 + (\frac{-2}{7}) + \frac{3}{8} + (\frac{-5}{7}) + 1\frac{2}{3} \\ =\frac{625}{1000} + (\frac{-2}{7}) + \frac{3}{8} + (\frac{-5}{7}) + \frac{5}{3} \\ =\frac{5}{8} + (\frac{-2}{7}) + \frac{3}{8} + (\frac{-5}{7}) + \frac{5}{3} \\ = (\frac{5}{8}+\frac{3}{8}) + [\left(\frac{-2}{7}\right) + \left(\frac{-5}{7}\right)] + \frac{5}{3} \\ =1 + (-1) + \frac{5}{3} = \frac{5}{3}. \\ d) (-3).\left(\frac{-38}{21}\right).\left(\frac{-7}{6}\right).\left(\frac{-3}{19}\right) \\ =\frac{(-3).(-38).(-7).(-3)}{21.6.19} \\ =\frac{(-3).(-7).(-38).(-3)}{21.6.19} \\ =\frac{3.7.2.19.3}{3.7.2.3.19} = 1 \\ e\left)\right(\frac{11}{8}:\frac{22}{9}).\frac{8}{5} = \frac{11}{8}.\frac{9}{22}.\frac{8}{5} = \frac{11.9.8}{9.2.11.2.5} = \frac{2}{5}. \\ g) \left[\right(\frac{-4}{5}).\frac{5}{8}]:\left(\frac{-25}{12}\right) = \frac{(-4).5}{5.8}:\left(\frac{-25}{12}\right) \\ = \frac{-1}{2}:\left(\frac{-25}{12}\right) = \frac{-1}{2}.\frac{12}{-25} = \frac{6}{25} \end{gathered}$$