Question

Умоляю помогите решить пожалуйста не проходите мимо ​

Answer 1

Ответ:

Объяснение:

$\bf\\x^2-2\sqrt{x^2-24} =39\\\\t=\sqrt{x^2-24}\geq 0 ;~t^2=x^2-24;~x^2=t^2+24\\\\t^2+24-2t=39\\\\t^2-2t+24-39=0\\\\t^2-2t-15=0\\\\\po~Vieta:~t_1=-3<0;~ne~udovl.\\~~~~~~~~~~~~~t_2=5\\\\\sqrt{x^2-24}=5\\\\x^2-24=25\\\\x^2=49\\\\Otvet:x=\pm7$

$\displaystyle\sqrt{4-x} <x+2\\\\4-x\geq 0;~x\leq 4\\x+2\geq 0;~x\geq -2\\\\\left \{ {{-2\leq x\leq 4} \atop {4-x<x^2+4x+4}} \right. \\\\\\\left \{ {{x\in[-2;4]} \atop {x^2+5x>0} \right.; \left \{ {{x\in[-2;4]} \atop {x(x+5)>0} \right.\\\\\\\left \{ {{x\in[-2;4]} \atop {+++++(-5)------(0)++++>x}} \right. \\\\\\\left \{ {{x\in[-2;4]} \atop {x\in(-\infty;-5)\cup(0;+\infty)}} \right. \\\\\\Otvet:x\in(0;4]$