Question

8 КЛАСС , ДАЮ 40 БАЛЛОВ

Answer 1

Ответ:

$1)\ \ \dfrac{x+5}{x+7}+\dfrac{14}{x+14}\ \ \ \Rightarrow \ \ \ \left\{\begin{array}{l}x+7\ne 0\\x+14\ne 0\end{array}\right\ \ \left\{\begin{array}{l}x\ne -7\\x\ne -14\end{array}\right\\\\\\x\in (-\infty ;-14\ )\cup (-14;-7\ )\cup (-7\ ;+\infty )\\\\\\b)\ \ \dfrac{8}{|x|-17}\ \ \ \Rightarrow \ \ \ |x|-17\ne 0\ \ ,\ \ \ |x|\ne 17\ \ ,\ \ \ x\ne \pm 17\\\\\\x\in (-\infty ;-17)\cup (-17;17\ )\cup (17;+\infty )$

$2)\ \ \dfrac{24a^{12}c^6}{36a^5c^{11}}=\dfrac{12\cdot 2\cdot a^5\cdot a^7\cdot c^6}{12\cdot 3\cdot a^5\cdot c^6\cdot c^5}=\dfrac{2\, a^7}{3\, c^5}$

$\displaystyle \frac{49-n^2}{n^2-14n+49}=\frac{(7-n)(7+n)}{(7-n)^2}=\frac{7+n}{7-n}\\\\\\\frac{x^3+64}{x^2-7xy+4x-28y}=\frac{(x+4)(x^2-4x+16)}{x(x-7y)+4(x-7y)}=\frac{(x+4)(x^2-4x+16)}{(x-7y)(x+4)}=\\\\=\frac{x^2-4x+16}{x-7y}$