$\frac{1}{a^2-ab}:\frac{b}{b^2-a^2}=\frac{1}{a(a-b)}:\frac{b}{(b-a)(b+a)}=\frac{1}{a(a-b)}:(-\frac{b}{(a-b)(a+b)})=\frac{1}{a(a-b)}\ast (-\frac{(a-b)(a+b)}{b})=\frac{1}{a}\ast (-\frac{a+b}{b})=-\frac{a+b}{ab}$
при $a=2\frac{1}{3},b=-\frac{3}{7}$:
$-\frac{2\frac{1}{3}-\frac{3}{7}}{2\frac{1}{3}\ast (-\frac{3}{7})}=\frac{\frac{7}{3}-\frac{3}{7}}{\frac{7}{3}\ast \frac{3}{7}}=\frac{\frac{49-9}{21}}{\frac{1}{1}\ast \frac{1}{1}}=\frac{40}{21}=1\frac{19}{21}$

$\frac{a^2+4ab+4b^2}{a^2-9b^2}:\frac{3a+6b}{2a-6b}=\frac{(a+2b{)}^2}{(a-3b)(a+3b)}:\frac{3(a+2b)}{2(a-3b)}=\frac{(a+2b{)}^2}{(a-3b)(a+3b)}\ast \frac{2(a-3b)}{3(a+2b)}=\frac{a+2b}{a+3b}\ast \frac{2}{3}=\frac{2(a+2b)}{3(a+3b)}$
при a = 4, b = −5:
$\frac{2(a+2b)}{3(a+3b)}=\frac{2(4+2\ast (-5))}{3(4+3\ast (-5))}=\frac{2(4-10)}{3(4-15)}=\frac{2\ast (-6)}{3\ast (-11)}=\frac{2\ast (-2)}{-11}=\frac{-4}{-11}=\frac{4}{11}$