$\frac{a}{b}$
при a = 2, b = 1:
$\frac{2}{1}=2$
при a = −1, b = −3:
$\frac{-1}{-3}=\frac{1}{3}$
при $a=\frac{1}{2}$, b = 0,2:
$\frac{\frac{1}{2}}{0,2}=\frac{\frac{1}{2}}{\frac{1}{5}}=\frac{1}{2}\ast \frac{5}{1}=2,5$
при a = 0,4, $b=-\frac{1}{3}$:
$\frac{0,4}{-\frac{1}{3}}=\frac{2}{5}\ast (-\frac{3}{1})=-\frac{6}{5}=-1,2$
 
$a-\frac{1}{b}$
при a = 2, b = 1:
$2-\frac{1}{1}=2-1=2$
при a = −1, b = −3:
$-1-\frac{1}{-3}=-1+\frac{1}{3}=-\frac{2}{3}$
при $a=\frac{1}{2}$, b = 0,2:
$\frac{1}{2}-\frac{1}{0,2}=\frac{1}{2}-\frac{1}{\frac{1}{5}}=\frac{1}{2}-5=-4,5$
при a = 0,4, $b=-\frac{1}{3}$:
$0,4-\frac{1}{-\frac{1}{3}}=0,4+3=3,4$
 
$\frac{a+b}{a}$
при a = 2, b = 1:
$\frac{2+1}{2}=\frac{3}{2}=1,5$
при a = −1, b = −3:
$\frac{-1-3}{-1}=\frac{-4}{-1}=4$
при $a=\frac{1}{2}$, b = 0,2:
$\frac{\frac{1}{2}+0,2}{\frac{1}{2}}=\frac{\frac{1}{2}+\frac{1}{5}}{\frac{1}{2}}=\frac{\frac{5+2}{10}}{\frac{1}{2}}=\frac{7}{10}\ast 2=\frac{7}{5}=1,4$
при a = 0,4, $b=-\frac{1}{3}$:
$\frac{0,4-\frac{1}{3}}{0,4}=\frac{\frac{2}{5}-\frac{1}{3}}{\frac{2}{5}}=\frac{\frac{6-5}{15}}{\frac{2}{5}}=\frac{1}{15}\ast \frac{5}{2}=\frac{1}{3}\ast \frac{1}{2}=\frac{1}{6}$
 
$\frac{a-b}{a+b}$
при a = 2, b = 1:
$\frac{2-1}{2+1}=\frac{1}{3}$
при a = −1, b = −3:
$\frac{-1-(-3)}{-1-3}=\frac{-1+3}{-4}=-\frac{2}{4}=-0,5$
при $a=\frac{1}{2}$, b = 0,2:
$\frac{\frac{1}{2}-0,2}{\frac{1}{2}+0,2}=\frac{0,5-0,2}{0,5+0,2}=\frac{0,3}{0,7}=\frac{3}{10}\ast \frac{10}{7}=\frac{3}{7}$
при a = 0,4, $b=-\frac{1}{3}$:
$\frac{0,4-(-\frac{1}{3})}{0,4-\frac{1}{3}}=\frac{\frac{2}{5}+\frac{1}{3}}{\frac{2}{5}-\frac{1}{3}}=\frac{\frac{6+5}{15}}{\frac{6-5}{15}}=\frac{11}{15}\ast \frac{15}{1}=11$
 
$\frac{a^2-b^2}{a-2b}$
при a = 2, b = 1:
$\frac{2^2-1^2}{2-2\ast 1}=\frac{4-1}{0}$ − не имеет смысла
при a = −1, b = −3:
$\frac{(-1{)}^2-(-3{)}^2}{-1-2\ast (-3)}=\frac{1-9}{-1+6}=\frac{-8}{5}=-1,6$
при $a=\frac{1}{2}$, b = 0,2:
$\frac{(\frac{1}{2}{)}^2-0,2^2}{\frac{1}{2}-2\ast 0,2}=\frac{\frac{1}{4}-0,04}{0,5-0,4}=\frac{0,25-0,04}{0,1}=\frac{0,21}{0,1}=2,1$
при a = 0,4, $b=-\frac{1}{3}$:
$\frac{0,4^2-(-\frac{1}{3}{)}^2}{0,4-2\ast (-\frac{1}{3})}=\frac{0,16-\frac{1}{9}}{0,4+\frac{2}{3}}=\frac{\frac{4}{25}-\frac{1}{9}}{\frac{2}{5}+\frac{2}{3}}=\frac{\frac{36-25}{225}}{\frac{6+10}{15}}=\frac{11}{225}\ast \frac{15}{16}=\frac{11}{15}\ast \frac{1}{16}=\frac{11}{240}$