Алгебра: Знайти область визначення функії

Знайти область визначення функії

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Ответ:

tka4ukilia2017

19.08.2021 20:32

$1)\ \ y=\dfrac{\sqrt{2x-3}}{x^2-3x+2}\\\\\\\left\{\begin{array}{l}2x-3\geq 0\\x^2-3x+2\ne 0\end{array}\right\ \ \left\{\begin{array}{l}x\geq 1,5\\x\ne 1\ ,\ x\ne 2\end{array}\right\ \ \ \Rightarrow \ \ \ x\in [\ 1,5\ ;2)\cup (\ 2\ ;+\infty \, )\\\\\\2)\ \ y=\dfrac{\sqrt{x^2-3x-4}}{x-4}$

$\left\{\begin{array}{l}x^2-3x-4\geq 0\\x-4\ne 0\end{array}\right\ \ \left\{\begin{array}{l}(x+1)(x-4)\geq 0\\x\ne 4\end{array}\right\ \ \left\{\begin{array}{l}x\in (-\infty ;-1\ ]\cup [\ 4\ ;+\infty )\\x\ne 4\end{array}\right\\\\\\x\in x\in (-\infty ;-1\ ]\cup (\ 4\ ;+\infty )$

$3)\ \ y=\dfrac{\sqrt{9-x^2}}{2x-1}\\\\\\\left\{\begin{array}{l}9-x^2\geq 0\\2x-1\ne 0\end{array}\right\ \ \left\{\begin{array}{l}(3-x)(3+x)\geq 0\\x\ne 0,5\end{array}\right\ \ \left\{\begin{array}{l}(x-3)(3+x)\leq 0\\x\ne 0,5\end{array}\right\\\\\\\left\{\begin{array}{l}x\in [-3\ ;\ 3\ ]\\x\ne 0,5\end{array}\right\ \ \ \Rightarrow \ \ \ x\in [-3\ ;0,5\ )\cup (\ 0,5\ ;\ 3\ ]$

$4)\ \ y=\dfrac{\sqrt{x^2-x-6}}{x-7}\\\\\\\left\{\begin{array}{l}x^2-x-6\geq 0\\x-7\ne 0\end{array}\right\ \ \left\{\begin{array}{l}(x+2)(x-3)\geq 0\\x\ne 7\end{array}\right\ \ \left\{\begin{array}{l}x\in (-\infty ;-2\ ]\cup [\ 3\ ;+\infty )\\x\ne 7\end{array}\right\\\\\\x\in (-\infty ;-2\ ]\cup [\ 3\ ;\ 7\ )\cup (\ 7\ ;+\infty )$