Под цифрой 2 и 9. Решите п

$\displaystyle \tt 2). \ \bigg(\frac{5x^{2}-15xy}{x^{2}-9y^{2}}-\frac{3xy+9y^{2}}{x^{2}+6xy+9y^{2}}\bigg):\frac{4y-18}{y^{2}-9}=\\\\\\ = \ \bigg(\frac{5x(x-3y)}{(x-3y)(x+3y)}-\frac{3y(x+3y)}{(x+3y)^{2}}\bigg)\cdot\frac{(y-3)(y+3)}{4y-18}=\\\\\\= \ \frac{5x-3y}{x+3y}\cdot\frac{(y-3)(y+3)}{4y-18}=\frac{(5x-3y)(y-3)(y+3)}{2(x+3y)(2y-9)};$

$\displaystyle \tt 9). \ 7\cdot\bigg(\frac{2}{3}\bigg)^{5}-\frac{2}{3}\cdot\bigg(\frac{4}{9}\bigg)^{2}+3\cdot\bigg(\frac{8}{27}\bigg)^{3}:\bigg(\frac{2}{3}\bigg)^{4}=\\\\\\= \ 7\cdot\bigg(\frac{2}{3}\bigg)^{5}-\frac{2}{3}\cdot\bigg(\frac{2}{3}\bigg)^{4}+3\cdot\bigg(\frac{2}{3}\bigg)^{9}:\bigg(\frac{2}{3}\bigg)^{4}=\\\\\\= \ 7\cdot\bigg(\frac{2}{3}\bigg)^{5}-\bigg(\frac{2}{3}\bigg)^{5}+3\cdot\bigg(\frac{2}{3}\bigg)^{5}=\bigg(\frac{2}{3}\bigg)^{5}\cdot(7-1+3)=$

$\displaystyle \tt = \ \frac{32\cdot9}{243}=\frac{32}{27}=1\frac{5}{27}$