Bài 1 trang 24 toán lớp 7 tập 1 CTSTBỏ dấu ngoặc rồi tính:a) $\left ( \frac{-3}{7}\right ) + \left...

a)
$\left ( \frac{-3}{7}\right ) + \left ( \frac{5}{6} -\frac{4}{7} \right )$
= $\left ( \frac{-3}{7}\right ) + \left ( \frac{35}{42} -\frac{24}{42} \right )$
= $\left ( \frac{-3}{7}\right ) + \frac{11}{42}$
= $\left ( \frac{-18}{42}\right ) + \frac{11}{42}$
= $\frac{-1}{6}$

b)
$\frac{3}{5} - \left (\frac{2}{3} +\frac{1}{5} \right )$
= $\frac{3}{5} - \left (\frac{10}{15} +\frac{3}{15} \right )$
= $\frac{3}{5} - \frac{13}{15} $
= $\frac{9}{15} - \frac{13}{15} $
= $\frac{-4}{15}$

c)
$\left [ \left ( \frac{-1}{3}\right ) + 1 \right ] - \left ( \frac{2}{3} -\frac{1}{5}\right )$
= $\left [ \left ( \frac{-1}{3}\right ) + \frac{3}{3} \right ] - \left ( \frac{10}{15} - \frac{3}{15}\right )$
= $\frac{2}{3} - \frac{7}{15}$
= $\frac{10}{15} - \frac{7}{15}$
= $\frac{1}{5}$

d)
$1\frac{1}{3} + \left (\frac{2}{3} -\frac{3}{4} \right ) - \left ( 0.8 + 1\frac{1}{5} \right )$
= $\frac{4}{3} + \left (\frac{8}{12} -\frac{9}{12} \right ) - \left ( \frac{4}{5} + \frac{6}{5} \right )$
= $\frac{4}{3} - \frac{1}{12} - 2$
= $\frac{16}{12} - \frac{1}{12} - \frac{24}{12}$
= $-\frac{9}{12}$
= $-\frac{3}{4}$