Решите срочно 15 задание б) г) е)

Решите задачу:

$1)\; \; f(x)=2x+ \sqrt{x^2-4x-12} \; ;\\\\x^2-4x-12 \geq 0\; \; \to \; \; (x+2)(x-6) \geq 0\\\\+++[-2\, ]---[\, 6\, ]+++\\\\x\in (-\infty ,-2\, ]\cup [\, 6,+\infty )\\\\2)\; \; f(x)= \sqrt{-3x^2+6x-18} + \frac{1}{x-8} \; ;\\\\ \left \{ {{-3x^2+6x-18 \geq 0|:(-3)} \atop {x-8\ne 0}} \right. \; \left \{ {{x^2-2x+6 \leq 0} \atop {x\ne 8}} \right. \; \left \{ {{x\in \varnothing } \atop {x\ne 8}} \right. \; \; \to \; \; x\in \varnothing \\\\P.S.\; x^2-2x+6=0\; ,\; \; D=-20\ \textless \ 0\; \; \Rightarrow \; \; x^2-2x+6 \ \textgreater \ 0$

$3)\; \; f(x)= \sqrt{-x^2+4x+32}+ \frac{1}{x^2-9} \; ;\\\\ \left \{ {{-x^2+4x+32 \geq 0} \atop {x^2-9\ne 0}} \right. \; \left \{ {{x^2-4x-32 \leq 0} \atop {(x-3)(x+3)\ne 0}} \right. \; \left \{ {{(x+4)(x-8) \leq 0} \atop {x\ne -3,\; x\ne 3}} \right. \; \left \{ {{x\in [-4,8\, ]} \atop {x\ne -3,\; x\ne 3}} \right. \\\\x\in (-4,-3)\cup (-3,3)\cup (3,8\, ]$