Вариант 3: 1) $\frac{16x^6y}{20x^2y^3}$; 2) $\frac{4x-16}{x^2-16}$; 3) $\frac{3y-9}{y^2-6y+9}$; 4) $\frac{c^3+c^5}{c^7+c^5}$; 5) $\frac{10y^2-15y}{3-2y}$; 6) $\frac{y^2-12y+36}{36-y^2}$; 7) $\frac{x^3-2x^2-3x+6}{x^3-8}$.

Question

Вопрос: $Фото задания: Вариант 3: 1) $\frac{16x^6y}{20x^2y^3}$; 2) $\frac{4x-16}{x^2-16}$; 3) $\frac{3y-9}{y^2-6y+9}$; 4)…$ Фото задания

Accepted Answer

$\frac{16x^6y}{20x^2y^3}=\frac{4x^4}{5y^2}$.
$\frac{4x-16}{x^2-16}=\frac{4(x-4)}{(x-4)(x+4)}=\frac4{x+4}$.
$\frac{3y-9}{y^2-6y+9}=\frac{3(y-3)}{(y-3)^2}=\frac3{y-3}$.
$\frac{c^3+c^5}{c^7+c^5}=\frac{c^3(c^2+1)}{c^5(c^2+1)}=\frac1{c^2}$.
$\frac{10y^2-15y}{3-2y}=\frac{5y(2y-3)}{-(2y-3)}=-5y$.
$\frac{y^2-12y+36}{36-y^2}=\frac{(y-6)^2}{(6-y)(6+y)}=\frac{6-y}{y+6}$.
$\frac{x^3-2x^2-3x+6}{x^3-8}=\frac{(x-2)(x^2-3)}{(x-2)(x^2+2x+4)}=\frac{x^2-3}{x^2+2x+4}$.

Ответ: 1) $\frac{4x^4}{5y^2}$; 2) $\frac4{x+4}$; 3) $\frac3{y-3}$; 4) $\frac1{c^2}$; 5) $-5y$; 6) $\frac{6-y}{y+6}$; 7) $\frac{x^2-3}{x^2+2x+4}$.

Вариант 3: 1) \(\frac{16x^6y}{20x^2y^3}\); 2) \(\frac{4x-16}{x^2-16}\); 3) \(\frac{3y-9}{y^2-6y+9}\); 4) \(\frac{c^3+c^5}{c^7+c^5}\); 5) \(\frac{10y^2-15y}{3-2y}\); 6) \(\frac{y^2-12y+36}{36-y^2}\); 7) \(\frac{x^3-2x^2-3x+6}{x^3-8}\).