Bài 3 (Trang 16, SGK Toán 7, Tập 1, Bộ Chân trời sáng tạo):

Thay ? bằng dấu (>, <, =) thích hợp.

a) $\left(\frac{-5}{8}\right) +\hspace{0.17em}\left(\frac{3}{-8}\right) \mathbf{?} -1;$

b) $\left(\frac{-13}{22}\right) + \left(\frac{-5}{22}\right) \mathbf{?} \frac{-8}{11};$

c) $\frac{1}{6} + \left(\frac{-3}{4}\right) \mathbf{?} \frac{1}{14} + \left(\frac{-4}{7}\right).$

Hướng dẫn giải:

$\mathrm{a}) \left(\frac{-5}{8}\right) + \left(\frac{3}{-8}\right) = \left(\frac{-5}{8}\right) + \left(\frac{-3}{8}\right) = \frac{-8}{8} = -1$

Vậy ta điền $\left(\frac{-5}{8}\right) +\hspace{0.17em}\left(\frac{3}{-8}\right) \mathbf{=}-1;$

b) $\left(\frac{-13}{22}\right) + \left(\frac{-5}{22}\right) = \frac{-18}{22} = \frac{-9}{11} < \frac{-8}{11}$

Vậy ta điền $\left(\frac{-13}{22}\right) + \left(\frac{-5}{22}\right) < \frac{-8}{11}$

c) $\frac{1}{6} + \left(\frac{-3}{4}\right) = \frac{2}{12} + \left(\frac{-9}{12}\right) = \frac{-7}{12}$

$\frac{1}{14} + \left(\frac{-4}{7}\right) = \frac{1}{14} + \left(\frac{-8}{14}\right) = \frac{-7}{14}$

Ta có $\frac{-7}{12} < \frac{-7}{14}$

Vậy ta điền $\frac{1}{6} + \left(\frac{-3}{4}\right) < \frac{1}{14} + \left(\frac{-4}{7}\right)$.