Question

помогите найти о.о.ф.
а) y=5/√x-5
в) y= $sqrt{x} 16- x^{2}$

Answer 1

$y= frac{5}{ sqrt{x-5} } \ \ left { {{x-5 geq 0} atop {x-5neq 0}} right. \ x-5>0 \ x>5$
D(y)=(5;+∞)

$y= sqrt{16-x^2}$
$16-x^2 geq 0 \ x^2 leq 16 \ -4 leq x leq 4$

D(y)=[-4;4]

Answer 2

спасибо