Question

Знайдіть область визначення функції:

Answer 1

Ответ:

$y=lg\dfrac{6}{2x^2-3x+1}\ \ \ ,\ \ \ \ OOF:\ \left\{\begin{array}{l}\dfrac{1}{2x^2-3x+1}>0\\2x^2-3x+1\ne 0 \end{array}\right\ \ \left\{\begin{array}{l}2x^2-3x+1>0\\2x^2-3x+1\ne 0\end{array}\right\\\\\\\left\{\begin{array}{l}2\, (x-\dfrac{1}{2})(x-1)>0\\2(x-\dfrac{1}{2})(x-1)\ne 0\end{array}\right\ \ \Rightarrow \ \ \ 2\, (x-\dfrac{1}{2})(x-1)>0\ \ ,\\\\\\znaki:\ \ +++(\dfrac{1}{2})---(1)+++\\\\x\in (-\infty \, ;\dfrac{1}{2}\, )\cup (\, 1\, ;+\infty \, )$