Вариант 2: 1) (12a⁸c)/(15a²c⁵); 2) (25−x²)/(25−5x); 3) (x²−10x+25)/(5x−25); 4) (a²+a⁵)/(a²+a⁵); 5) (12y²−8y)/(2−3y); 6) (a²−9)/(9−6a+a²); 7) (x³+8)/(x³+2x²−3x−6).

Question

Вопрос:

Фото задания: Вариант 2: 1) (12a⁸c)/(15a²c⁵); 2) (25−x²)/(25−5x); 3) (x²−10x+25)/(5x−25); 4) (a²+a⁵)/(a²+a⁵); 5) (12y²−8y)/(2−3y); 6)…

Вариант 2: 1) (12a⁸c)/(15a²c⁵); 2) (25−x²)/(25−5x); 3) (x²−10x+25)/(5x−25); 4) (a²+a⁵)/(a²+a⁵); 5) (12y²−8y)/(2−3y); 6) (a²−9)/(9−6a+a²); 7) (x³+8)/(x³+2x²−3x−6).

Accepted Answer

\(\frac{12a^8c}{15a^2c^5}=\frac{4a^6}{5c^4}\).
\(\frac{25-x^2}{25-5x}=\frac{(5-x)(5+x)}{5(5-x)}=\frac{x+5}{5}\).
\(\frac{x^2-10x+25}{5x-25}=\frac{(x-5)^2}{5(x-5)}=\frac{x-5}{5}\).
\(\frac{a^2+a^5}{a^2+a^5}=1\).
\(\frac{12y^2-8y}{2-3y}=\frac{4y(3y-2)}{-(3y-2)}=-4y\).
\(\frac{a^2-9}{9-6a+a^2}=\frac{(a-3)(a+3)}{(a-3)^2}=\frac{a+3}{a-3}\).
\(\frac{x^3+8}{x^3+2x^2-3x-6}=\frac{(x+2)(x^2-2x+4)}{(x+2)(x^2-3)}=\frac{x^2-2x+4}{x^2-3}\).

Ответ: 1) \(\frac{4a^6}{5c^4}\); 2) \(\frac{x+5}{5}\); 3) \(\frac{x-5}{5}\); 4) \(1\); 5) \(-4y\); 6) \(\frac{a+3}{a-3}\); 7) \(\frac{x^2-2x+4}{x^2-3}\).