Ответ :

[tex]y= \frac{3}{ x^{2} +9} [/tex] x∈R
[tex]y= \frac{5x-15}{x(x-3)} [/tex]  x≠0  x≠3
[tex]y = \frac{ \sqrt{2x+1} }{ \sqrt{x-3} } [/tex]
[tex] \left \{ {{2x+1 \geq 0} \atop {x-3 \geq>0}} \right. [/tex]
[tex] \left \{ {{x \geq - \frac{1}{2} } \atop {x>3}} \right. [/tex]
⇒ x>3
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[tex]y= \frac{ \sqrt{2x+1} }{x-3} [/tex]
[tex] \left \{ {{x \geq - \frac{1}{2} } \atop {x \neq 3}} \right. [/tex]
x∈[[tex]- \frac{1}{2} ;3) (3; \infty) [/tex]