найти область определения функцииy=3/x^2+9y=5x-15/x(x-3)y=√2x+1/√x-3y=√2x+1/x-3

$y= \frac{3}{ x^{2} +9}$ x∈R
$y= \frac{5x-15}{x(x-3)}$ x≠0 x≠3
$y = \frac{ \sqrt{2x+1} }{ \sqrt{x-3} }$

0}} \right. " alt=" \left \{ {{2x+1 \geq 0} \atop {x-3 \geq>0}} \right. " align="absmiddle" class="latex-formula">

3}} \right. " alt=" \left \{ {{x \geq - \frac{1}{2} } \atop {x>3}} \right. " align="absmiddle" class="latex-formula">
⇒ x>3
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$y= \frac{ \sqrt{2x+1} }{x-3}$
$\left \{ {{x \geq - \frac{1}{2} } \atop {x \neq 3}} \right.$
x∈[ $- \frac{1}{2} ;3) (3; \infty)$