Второй вариант хоть что нибудь

Решите задачу:

$1)\quad y= \frac{\sqrt{8+6x+x^2}}{|x+4|-3}\\\\OOF:\; \; \left \{ {{x^2+6x+8 \geq 0} \atop {|x+4|-3\ne 0}} \right. \; \left \{ {{(x+2)(x+4) \geq 0} \atop {|x+4|\ne 3}} \right. \; \left \{ {{(x+2)(x+4) \geq 0} \atop {x+4\ne \pm 3}} \right. \\\\\left \{ {{x\in (-\infty ,-4)\cup (-2,+\infty )} \atop {x\ne -1,\; x\ne -7}} \right. \\\\x\in (-\infty ,-7)\cup (-7,-4\, ]\cup [-2,+\infty )$

$2)\; \; y=\frac{4}{3+\sqrt{|x-5|+1}}\\\\|x-5| \geq 0\; \; \to \; \; |x-5|_{min}=0\; \; \to \; \; (|x-5|+1)_{min}=1\\\\y_{min}= \frac{4}{3+\sqrt1} =\frac{4}{3+1}=1\\\\pri\; \; x\to \infty :\; \; |x-5|\to \infty \; ,\; \; \sqrt{|x-5|+1}\to \infty\; ,\\\\(3+\sqrt{|x-5|+1} )\to \infty \; \; \; \Rightarrow \; \; \; \frac{4}{\infty }\to 0\; \; \Rightarrow \; \; y\to 0\\\\y\in (0,1\, ]$

$3)\; \; y=4x^4-2|x|\cdot x^2\\\\y(-x)=4(-x)^4-2\cdot |-x|\cdot (-x)^2=4x^4-2\cdot |x|\cdot x^2=\\\\=4x^4-2|x|\cdot x^2=y(x)\\\\y(-x)=y(x)\; \; \to \; \; \; y(x)\; \; - \; chetnaya\\\\P,S,\; \; \; |-x|=|-1\cdot x|=|-1|\cdot |x|=1\cdot |x|=|x|;$