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

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