Question

Выполнить действия: а) (х^2-64)/(х^2+8х+16)·(2х+8)/〖(х-8)〗^2 ;
b)(3а-6b)/〖(a-2b)〗^2∶(3(2b-a))/(4a-8b);
с) (5m/(3n^2 k))^2 ∶15m^4 k^3

Answer 1

Объяснение:

$\frac{x^{2} -64}{x^{2} +8x+16} * \frac{2x+8}{(x-8)^{2}} =$  $\frac{(x+8)(x-8)}{(x+4)(x+4)} * \frac{2(x+4)}{(x-8)(x-8)} =$$\frac{2(x+8)}{(x+4)(x-8)}$

$\frac{3a-6b}{(a-2b)^{2}} : \frac{3(2b-a)}{4a-8b} =$$\frac{3(a-2b)}{(a-2b)(a-2b)} * \frac{4(a-2b)}{(-3)(a-2b)} =$$\frac{4}{2b-a}$

$(\frac{5m}{3n^{2}k} )^{2} * \frac{1}{15m^{4}k^{3} } =$$\frac{25m^{2}}{9n^{4}k^{2} } * \frac{1}{15m^{4}k^{3} } = \frac{5}{27k^{5}n^{4}m^{2} }$