Question

Найдите область определения выражения:​

Answer 1

$1)\ \ \sqrt{4-3x}\ \ \to \ \ \ 4-3x\geq 0\ \ ,\ \ \ -3x\geq -4\ \ ,\ \ \ 3x\leq 4\ \ ,\ \ x\leq \dfrac{4}{3}\\\\x\in (-\infty \, ;\ 1\frac{1}{3}\ ]\\\\\\2)\ \ \dfrac{1}{\sqrt{y-5}}\ \ \to \ \ \ y-5>0\ \ ,\ \ \ y>5\ \ ,\ \ \ y\in (\, 5\, ;+\infty \, )$

$3)\ \ \dfrac{1}{\sqrt{y}-5}\ \ \to \ \ \ \left\{\begin{array}{l}\sqrt{y}-5\ne 0\\y\geq 0\end{array}\right\ \ \left\{\begin{array}{l}\sqrt{y}\ne 5\\y\geq 0\end{array}\right\ \ \left\{\begin{array}{l}y\ne 25\\y\geq 0\end{array}\right\ \ \to \\\\\\y\in [\ 0\, ;\, 25\, )\cup (\, 25\, ;+\infty \, )\\\\\\4)\ \ \dfrac{1}{\sqrt{y}+5}\ \ \to \ \ \ \left\{\begin{array}{l}\sqrt{y}+5\ne 0\\y\geq 0\end{array}\right\ \ \left\{\begin{array}{l}\sqrt{y}\ne -5\\y\geq 0\end{array}\right\ \ \ \to \ \ \ y\geq 0$

$y\in [\ 0\, ;\, +\infty \, )$