Question

определите область определения функции

Answer 1

a)
$y=frac{1}{x^{2}+x} \ 1geq 0 \ x^{2}+xneq 0 \ x(x+1)neq 0\ xneq0 xneq -1$
x=1

a)
$y=frac{x^{2}-4}{sqrt{x-1}} \ x^{2}-4geq 0\ sqrt{x-1}neq 0 \ xgeq +-2 \ xneq 1$
x=[-2;2]

a)
$y=frac{sqrt{20-x-x^{2}}}{4-x^{2}} \ sqrt{20-x-x^{2}}geq 0\ 4-x^{2}neq 0\ x^{2}+x-20leq 0\ x_{1}=5 x_{2}=-4 \ xneq 2$
x=[-4;5]