0)} \atop {x\ne 3\qquad \qquad \qquad }} \right. \\\\\underline {x\in (-\infty ,-2)\cup (-2,3)\cup (3,+\infty )}\\\\2)\; \; f(x)=\frac{1}{x-3}\; \; \Rightarrow \; \; x-3\ne 0\; ,\; x\ne 3\\\\\underline {x\in (-\infty ,3)\cup (3,+\infty )}\\\\3)\; \; f(x)=x^2-1\; \; \Rightarrow \; \; \underline {x\in (-\infty ,+\infty )}" alt=" 1)\; \; f(x)=\frac{x}{(x^3+8)(x-3)^2}\; \; \Rightarrow \; \; (x^3+8)(x-3)^2\ne 0\\\\\left \{ {{x^3+8\ne 0} \atop {x-3\ne 0}} \right. \; \left \{ {{(x+2)(x^2-2x+4)\ne 0} \atop {x\ne 3}} \right. \; \left \{ {{x\ne -2\; \; (x^2-2x+4>0)} \atop {x\ne 3\qquad \qquad \qquad }} \right. \\\\\underline {x\in (-\infty ,-2)\cup (-2,3)\cup (3,+\infty )}\\\\2)\; \; f(x)=\frac{1}{x-3}\; \; \Rightarrow \; \; x-3\ne 0\; ,\; x\ne 3\\\\\underline {x\in (-\infty ,3)\cup (3,+\infty )}\\\\3)\; \; f(x)=x^2-1\; \; \Rightarrow \; \; \underline {x\in (-\infty ,+\infty )}" align="absmiddle" class="latex-formula">