[tex]a)\\
(3\frac12+1\frac12)*\frac35=5*\frac35=3\\
b)\\
(2\frac23-\frac45)*1\frac37=(2\frac{10}{15}-\frac{12}{15})*\frac{10}7=(1\frac{25}{15}-\frac{12}{15})*\frac{10}7=1\frac{13}{15}*\frac{10}7=\frac{28}{15}*\frac{10}7=\frac{4}{3}*\frac{2}{1}=\frac83=2\frac23[/tex]
[tex]c)\\
(15\frac23-7\frac12):1\frac12=(15\frac{4}{6}-7\frac36):\frac32=8\frac16:\frac32=\frac{49}{6}*\frac23=\frac{49}{3}*\frac{1}{3}=\frac{49}3=16\frac13[/tex]
[tex]d)\\
2*(\frac13)^2+3*(\frac12)^3=2*\frac19+3*\frac18=\frac29+\frac38=\frac{16}{72}+\frac{27}{72}=\frac{43}{72}[/tex]
[tex]e)\\
4\frac12*1\frac13-2\frac25*\frac57=\frac92*\frac43-\frac{12}5*\frac57=\frac{3}{1}*\frac{2}{1}-\frac{12}{1}*\frac{1}{7}=6-\frac{12}7=4\frac{14}{7}-\frac{12}7=4\frac27[/tex]
[tex]f)\\
1\frac23-\frac56+2\frac1{10}:1\frac6{15}=\frac53-\frac56+\frac{21}{10}:\frac{21}{15}=\frac{10}{6}-\frac{5}6+\frac{21}{10}*\frac{15}{21}=\frac{10}6-\frac56+\frac{3}{2}=\frac{10}6-\frac56+\frac{9}{6}=\frac56+\frac96=\frac{14}6=2\frac26=2\frac13[/tex]
