Giải bài 6 trang 25 toán 7 tập 1 chân trời sáng tạo.
a) $\frac{13}{23}.\frac{7}{11}+\frac{10}{23}.\frac{7}{11}$
= $\frac{7}{11}. (\frac{13}{23}+\frac{10}{23})$
= $\frac{7}{11}. 1$
= $\frac{7}{11}$
b) $\frac{5}{9}.\frac{23}{11}-\frac{1}{11}.\frac{5}{9} + \frac{5}{9}$
= $\frac{5}{9} . (\frac{23}{11} - \frac{1}{11} + 1)$
= $\frac{5}{9} . (\frac{23}{11} - \frac{1}{11} + 1)$
= $\frac{5}{9} . (\frac{23}{11} - \frac{1}{11} + 1)$
= $\frac{5}{9} . 3$
= $\frac{5}{3}$
c) $\left [ \left ( -\frac{4}{9} \right ) +\frac{3}{5}\right ]: \frac{13}{17}+\left ( \frac{2}{5} -\frac{5}{9}\right ):\frac{13}{17}$
= $\left [ \left ( -\frac{4}{9} \right ) +\frac{3}{5} +\frac{2}{5} -\frac{5}{9} \right ] :\frac{13}{17}$
= $\left [ \left ( -\frac{4}{9} -\frac{5}{9}\right )+\left ( \frac{3}{5} +\frac{2}{5}\right ) \right ] :\frac{13}{17}$
= $(-1+1) :\frac{13}{17}$
= 0
d) $\frac{3}{16} : \left ( \frac{3}{22}-\frac{3}{11}\right )+\frac{3}{16}:\left ( \frac{1}{10}-\frac{2}{5} \right )$
= $\frac{3}{16} : \left ( \frac{3}{22}-\frac{3}{11}+ \frac{1}{10}-\frac{2}{5}\right )$
= $\frac{3}{16} : \left ( \frac{3}{22}-\frac{6}{22}+ \frac{1}{10}-\frac{4}{10}\right )$
= $\frac{3}{16} : \left ( -\frac{3}{22}-\frac{3}{10}\right )$
= $\frac{3}{16} : \left ( -\frac{3}{22}-\frac{3}{10}\right )$
= $\frac{3}{16} : \frac{24}{55}$
= $\frac{55}{128}$