Question

Знайдіть область визначення функції
$y=\frac{log_{10}(x^{2}-1 ) }{log_{2}(10-2x )}$

Answer 1

$\displaystyle\bf\\y=\frac{\log_{10}(x^{2} -1) }{\log_{2} (10-2x)} \\\\\\\left\{\begin{array}{ccc}x^{2} -1 > 0\\10-2x > 0\\\log_{2} (10-2x)\neq 0\end{array}\right\\\\\\\left\{\begin{array}{ccc}(x-1)(x+1) > 0\\-2x > -10\\10-2x\neq 1\end{array}\right\\\\\\\left\{\begin{array}{ccc}x\in(-\infty \ ; \ -1) \ \cup \ (1 \ ; \ +\infty)\\x < 5\\-2x\neq -9\end{array}\right \\\\\\\left\{\begin{array}{ccc}x\in(-\infty \ ; \ -1) \ \cup \ (1 \ ; \ +\infty)\\x < 5\\x\neq 4,5\end{array}\right$

$\displaystyle\bf\\Otvet \ : \ x\in(-\infty \ ; \ -1) \ \cup \ (1 \ ; \ 4,5) \ \cup \ (4,5 \ ; \ 5)$