Question

решите пожайлуста 2.14 полностью

Answer 1

$1)\; \; y=\sqrt{4-|x|}+\frac{1}{x+2}\; \; \Rightarrow \; \; \left \{ {{4-|x|\geq 0} \atop {x+2\ne 0}} \right.\; \; \left \{ {{|x|\leq 4} \atop {x\ne -2}} \right. \; \; \left \{ {{-4\leq x\leq 4} \atop {x\ne -2}} \right. \; \; \Rightarrow \\\\x\in [-4,-2)\cup (-2,4\, ]$

$2)\; \; y=\sqrt{|x|-3}+\frac{1}{\sqrt{x+1}}\; \; \Rightarrow \; \; \left \{ {{|x|-3\geq 0} \atop {x+1>0}} \right.\; \; \left \{ {{|x|\geq 3} \atop {x>-1}} \right. \; \; \left \{ {{x\geq 3\; \; ili\; \; x\leq -3} \atop {x>-1}} \right. \; \; \Rightarrow \\\\x\geq 3\; \; ,\; \; x\in [\, 3,+\infty )$

$3)\; \; y=\sqrt{|x+1|\, (x-3)}\; \; \Rightarrow \; \; |x+1|\, (x-3)\geq 0\; \; \Rightarrow \\\\\left \{ {{|x+1|\geq 0} \atop {x-3\geq 0}} \right. \; \; \left \{ {{|x+1|>0\; \; ili\; \; |x+1|=0} \atop {x\geq 3}} \right. \; \; \left \{ {{x\in R\; \; ili\; \; x=-1} \atop {x\geq 3}} \right. \; \; \; \Rightarrow \\\\x\in [\; 3\, ,+\infty )$