Решите по действиям Шестой класс математика

а)[latex](1 \frac{11}{18}+ \frac{13}{24}-1 \frac{13}{36} )*1 \frac{3}{5} =[/latex][latex]1 \frac{4}{15} [/latex] 1→[latex]1 \frac{11}{18}+ \frac{13}{24}= \frac{29}{18}+\frac{13}{24}= \frac{116}{72}+ \frac{39}{72}= \frac{155}{72} [/latex] 2→[latex] \frac{155}{72}-1 \frac{13}{36}=\frac{155}{72}-\frac{49}{72}= \frac{57}{72} [/latex] 3→[latex]\frac{57}{72} *1 \frac{3}{5}=\frac{57}{72} *\frac{8}{5}= \frac{57*8}{72*5}= \frac{57}{45}= 1 \frac{12}{45}=1 \frac{4}{15} [/latex] б)[latex] \frac{14}{15}*1 \frac{3}{7}+ \frac{7}{10}* \frac{3}{14}= [/latex][latex] 1 \frac{29}{60} [/latex] 1→[latex]\frac{14}{15}*1 \frac{3}{7}=\frac{14}{15}* \frac{10}{7}= \frac{14*10}{15*7}= \frac{4}{3} [/latex] 2→[latex]\frac{7}{10}* \frac{3}{14}= \frac{7*3}{10*14}= \frac{3}{20} [/latex] 3→[latex]\frac{4}{3}+\frac{3}{20}= \frac{80}{60} + \frac{9}{60}= \frac{89}{60}= 1 \frac{29}{60} [/latex] в)[latex]4 \frac{1}{6} * \frac{4}{15}- \frac{7}{18}*2 \frac{2}{3}= [/latex][latex] \frac{2}{27} [/latex] 1→[latex]4 \frac{1}{6} * \frac{4}{15}= \frac{25}{6}* \frac{4}{15}= \frac{25*4}{6*15}= \frac{10}{9} [/latex] 2→[latex]\frac{7}{18}*2 \frac{2}{3}=\frac{7}{18}* \frac{8}{3}= \frac{7*8}{18*3}= \frac{28}{27} [/latex] 3→[latex]\frac{10}{9}-\frac{28}{27}= \frac{30}{27}- \frac{28}{27}= \frac{2}{27} [/latex] г)[latex](1 \frac{13}{18}-2 \frac{7}{15}- \frac{4}{15} )* \frac{1}{2}= [/latex][latex]- \frac{91}{180} [/latex] 1→[latex]1 \frac{13}{18}-2 \frac{7}{15}=\frac{31}{18}-\frac{37}{15}= \frac{155}{90} - \frac{222}{90}=- \frac{67}{90} [/latex] 2→[latex]- \frac{67}{90}-\frac{4}{15}=- \frac{67}{90}- \frac{24}{90}=- \frac{91}{90} [/latex] 3→[latex]-\frac{91}{90}* \frac{1}{2}=- \frac{91*1}{90*2}=- \frac{91}{180} [/latex]