Question

Под цифрой 2 и 9. Решите п

Answer 1

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

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

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