Решите 65 и 66 номера с помощью замены 50 баллов

Решите задачу:

$1)\; \; \frac{5}{|x+2|+2} \ \textgreater \ |x+2|-2\\\\t=|x+2| \geq 0\; ,\; \; \frac{5}{t+2} \ \textgreater \ t-2\\\\ \frac{5}{t+2}-t+2 \ \textgreater \ 0\; ,\; \; \frac{5-t^2-2t+2t+4}{t+2} \ \textgreater \ 0\, |\cdot 9-1)\\\\ \frac{t^2-9}{t+2} \ \textless \ 0\; ,\; \frac{(t-3)(t+3)}{t+2} \ \textless \ 0\\\\Znaki:\; \; ---(-3)+++(-2)---(3)+++\; \; \left \{ {{t\ \textless \ -3} \atop {-2\ \textless \ t\ \textless \ 3}} \right. \\\\0\ \textless \ t\ \textless \ 3\; \; \to \; \; \left [ {{{|x+2|}\ \textgreater \ 0} \atop {|x+2|\ \textless \ 3}} \right. \; \; \to \; \; -3\ \textless \ x+2\ \textless \ 3\\\\-5\ \textless \ x\ \textless \ 1\\\\x\in (-5,1)$

$2)\; \; \frac{2}{|2x+1|+1} \leq -|2x+1|+2\\\\t=|2x+1| \geq 0,\; \; \frac{2}{t+1}+t-2 \leq 0\\\\ \frac{2+t^2+t-2t-2}{t+1} \leq 0\; ,\; \; \frac{t^2-t}{t+1} \leq 0\; ,\; \; \frac{t(t-1)}{t+1} \leq 0\\\\Znaki:\; ---(-1)+++(0)---(1)+++\; \left \{ {{t\ \textless \ -1} \atop {0 \leq t \leq 1}} \right. \\\\t \leq 1\; ,\; \; |2x+1| \leq 1\; \; \to \; \; -1 \leq 2x+1 \leq 1\\\\ -2\leq 2x \leq 0\; ,\; \; -1 \leq x \leq 0\\\\x\in [\, -1,0\, ]$