Đáp án:

 $\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ \frac{13}{35} ;\ b.\ \frac{5}{23} ;\ c.\ -4;\ d.\ 1\\ e.\ \frac{8}{19} ;\ f.\ \frac{5}{12} ;\ g.\ 1;\ h.\ \frac{160}{39}\\ i.\ \frac{14}{19} ;\ j.\ -1;\ k.\ \frac{5}{9} ;\ l.\ 13 \end{array}$

Giải thích các bước giải:

 $\displaystyle \begin{array}{{>{\displaystyle}l}} a.\ \frac{1}{7} .\frac{3}{15} +\frac{1}{7} .\frac{12}{5} =\frac{1}{7} .\left(\frac{3}{15} +\frac{12}{5}\right) =\frac{1}{7} .\left(\frac{3+12.3}{15}\right) =\frac{1}{7} .\frac{39}{15} =\frac{13}{35}\\ b.\ \frac{5}{23} .\frac{17}{26} +\frac{5}{23} .\frac{9}{26} =\frac{5}{23} .\left(\frac{17}{26} +\frac{9}{26}\right) =\frac{5}{23} .1=\frac{5}{23}\\ c.\ \frac{4}{9} .\frac{13}{3} -\frac{4}{3} .\frac{40}{9} =\frac{4}{9} .\left(\frac{13}{3} -\frac{40}{3}\right) =\frac{4}{9} .( -9) =-4\\ d.\ \frac{6}{7} +\frac{1}{7} .\frac{2}{7} +\frac{1}{7} .\frac{5}{7} =\frac{6}{7} +\frac{1}{7} .\left(\frac{2}{7} +\frac{5}{7}\right) =\frac{6}{7} +\frac{1}{7} .1=\frac{7}{7} =1\\ e.\ \frac{8}{19} .\frac{15}{7} -\frac{8}{19} .\frac{6}{7} -\frac{8}{19} .\frac{2}{7} =\frac{8}{19} .\left(\frac{15}{7} -\frac{6}{7} -\frac{2}{7}\right) =\frac{8}{19}\\ f.\ \frac{5}{12} .\frac{13}{17} -\frac{5}{12} .\frac{4}{17} +\frac{5}{12} .\frac{8}{17} =\frac{5}{12} .\left(\frac{13}{17} -\frac{4}{17} +\frac{8}{17}\right) =\frac{5}{12}\\ g.\ \frac{7}{19} .\frac{8}{11} +\frac{7}{19} .\frac{3}{11} +\frac{12}{19} =\frac{7}{19} .\left(\frac{8}{11} +\frac{3}{11}\right) +\frac{12}{19} =\frac{7}{19} +\frac{12}{19} =1\\ h.\ \frac{5}{9} .\frac{7}{13} +\frac{5}{9} .\frac{92}{13} -\frac{5}{9} .\frac{3}{13} =\frac{5}{9} .\left(\frac{7}{13} +\frac{92}{13} -\frac{3}{13}\right) =\frac{5}{9} .\frac{96}{13} =\frac{160}{39}\\ i.\ \frac{5}{19} .\frac{7}{13} +\frac{7}{19} .\frac{25}{13} -\frac{7.4}{19.13} =\frac{7}{13} .\left(\frac{5}{19} +\frac{25}{19} -\frac{4}{19}\right) =\frac{7}{13} .\frac{26}{19} =\frac{14}{19}\\ j.\ \frac{-4}{9} .\frac{14}{29} +\frac{-4}{9} .\frac{15}{29} +\frac{-5}{9} =\frac{-4}{9} .\left(\frac{14}{29} +\frac{15}{29}\right) +\frac{-5}{9} =\frac{-4}{9} +\frac{-5}{9} =-1\\ k.\ \frac{1}{7} .\frac{5}{9} +\frac{5}{9} .\frac{2}{7} +\frac{5}{9} .\frac{1}{7} +\frac{5}{9} .\frac{3}{7} =\frac{5}{9} .\left(\frac{1}{7} +\frac{2}{7} +\frac{1}{7} +\frac{3}{7}\right) =\frac{5}{9}\\ l.\ 12.\left(\frac{1}{4} +\frac{5}{6}\right) =12.\left(\frac{3+5.2}{12}\right) =13 \end{array}$