Question

найти область определения функции с русском пожалуйста ​

Answer 1

Объяснение:

219.

a)

$z=\sqrt{ln(x^2-4y)} \\\left \{ {{x^2-4y > 0} \atop {ln(x^2-4x)\geq 0}} \right.\ \ \ \ \ \left \{ {{4y < x^2\ |:4} \atop {x^2-4y\geq e^0 }} \right. \ \ \ \ \left \{ {{y < \frac{x^2}{4} } \atop {4y\leq x^2-1\ |:4}} \right. \ \ \ \ \left \{ {{y < \frac{x^2}{4} } \atop {y\leq \frac{x^2}{4}-\frac{1}{4} }} \right. .\ \ \ \ \ \Rightarrow$

ОДЗ: $y\leq \frac{x^2}{4}-\frac{1}{4} .$

б)

$\left\{\begin{array}{ccc}sin(arcsin\sqrt{x^2+y^2})=\sqrt{x^2+y^2} \leq 1\\z\neq 0\\\ x^2+y^2\geq 0 \end{array}\right\ \ \ \ \ \left\{\begin{array}{ccc}x^2+y^2\leq 1\\z\neq 0\\x^2+y^2\geq 0\end{array}\right .\ \ \ \ \ \Rightarrow$

ОДЗ: $0\leq x^2+y^2\leq 1 \ \ \ \ z\neq 0.$