Помогите решить систему уравнений!!!!!Плиз

Решаем 1-ое неравенство, а потом второе.
0\\ 49t^2-99t+2 \leq 0,\\ D=9409, \sqrt{D}=97\\1=-\frac{1}{49}<0 ,t_2=2,\to \\-\frac{1}{49} \leq t \leq 2" alt="1) 7^{x+2}+2\cdot 7^{-x} \leq 99\\49\cdot 7^x+2\cdot \frac{1}{7^x}-99 \leq 0, \to t=7^x>0\\ 49t^2-99t+2 \leq 0,\\ D=9409, \sqrt{D}=97\\1=-\frac{1}{49}<0 ,t_2=2,\to \\-\frac{1}{49} \leq t \leq 2" align="absmiddle" class="latex-formula">
0,x\ne 1\\\frac{-3}{1+y}-y-5 \geq 0\\\frac{-y^2-6y-8}{y+1} \geq 0\\\frac{-(y+4)(y+2)}{y+1} \geq 0, \frac{(y+4)(y+2)}{y+1} \leq 0\\\\ - - - - - - [-4]+++++++[-2]- - - - - -[-1]+++++ " alt="0<7^x \leq 2\\7^x \leq 7^{log_7{2}}, x \leq log_7{2}, log_7{2}=\frac{lg2}{lg7}=0,36\\2)log_{2x}0,125=log_{2x}{(\frac{1}{2})^3}=log_{2x}{2^{-3}}=-3log_{2x}{2}=\\=-3\frac{1}{log_2{2x}}=\frac{-3}{1+log_2x}\\log_2(64x)=log_2(2^6x)=6log_2{2}+log_2x=6+log_2x\\\frac{-3}{1+log_2x}-6-log_2x+1 \geq 0\\y=log_2x, x>0,x\ne 1\\\frac{-3}{1+y}-y-5 \geq 0\\\frac{-y^2-6y-8}{y+1} \geq 0\\\frac{-(y+4)(y+2)}{y+1} \geq 0, \frac{(y+4)(y+2)}{y+1} \leq 0\\\\ - - - - - - [-4]+++++++[-2]- - - - - -[-1]+++++ " align="absmiddle" class="latex-formula">
$y\in (-\infty,-4]U[-2,-1]\\log_2{x} \leq -4,log_2{x} \leq log_2{2^{-4}}, x \leq \frac{1}{16}\\ \left \{ {{log_2{x} \geq -2} \atop {log_2{x} \leq {-1}} \right. \left \{ {{x \leq \frac{1}{4}} \atop {x \leq \frac{1}{2}}} \right.\\x\in (-\infty,\frac{1}{16}]U[\frac{1}{4},\frac{1}{2}]$
Вывод:
$x\in [\frac{1}{4},\log_7{2}]$