Giải câu 2 trang 22 sách toán VNEN lớp 6 tập 2

a) $\frac{-4}{5} + \frac{1}{-5} \;...\; 1$;

Ta có: $\frac{-4}{5} + \frac{1}{-5} = \frac{-4}{5} + \frac{-1}{5} = \frac{-4\;+\;(-1)}{5} = \frac{-5}{5} = -1$.

Vậy: $\frac{-4}{5} + \frac{1}{-5} \;<\; 1$;

b) $\frac{-13}{22} + \frac{-7}{22} \;...\; \frac{-8}{11}$;

Ta có: $\frac{-13}{22} + \frac{-7}{22} = \frac{-13\;+\;(-7)}{22} = \frac{-20}{22} = \frac{-10}{11} < \frac{-8}{11}$.

Vậy: $\frac{-13}{22} + \frac{-7}{22} \;<\; \frac{-8}{11}$;

c) $\frac{3}{5} \;...\;  \frac{2}{3} + \frac{-1}{5}$;

Ta có: $\frac{2}{3} + \frac{-1}{5} = \frac{10}{15} + \frac{-3}{15} = \frac{10\;+\;(-3)}{15} = \frac{7}{15}$;

Mà: $\frac{3}{5} = \frac{9}{15} > \frac{7}{15}$ nên $\frac{3}{5} \;>\;  \frac{2}{3} + \frac{-1}{5}$;

d) $\frac{1}{6} + \frac{-3}{4}\;...\;  \frac{1}{14} + \frac{-4}{7}$.

Ta có: $\frac{1}{6} + \frac{-3}{4} = \frac{2}{12} + \frac{-9}{12} = \frac{2\;+\;(-9)}{12} = \frac{-7}{12}$;

$\frac{1}{14} + \frac{-4}{7} = \frac{1}{14} + \frac{-8}{14} =\frac{1\;+\;(-8)}{14} = \frac{-7}{14} < \frac{-7}{12}$.

Vậy: $\frac{1}{6} + \frac{-3}{4}\;>\;  \frac{1}{14} + \frac{-4}{7}$.