[m]\left\{\begin{matrix} x>0;x\neq 1\\ x>2; x\neq 3 \\(log_{x}(x-2)-log_{x-2}(x))(log_{x}(x-2)+log_{x-2}(x))\leq 0 \end{matrix}\right.[/m]
[m]log_{x}(x-2)=\frac{1}{log_{x-2}x}[/m]
[m]\left\{\begin{matrix} x>2\\ x\neq 3 \\(\frac{1}{log_{x-2}(x)}-log_{x-2}(x))(\frac{1}{log_{x-2}(x)}+log_{x-2}(x))\leq 0 \end{matrix}\right.[/m]
[m]\left\{\begin{matrix} x>2\\ x\neq 3 \\\frac{1-log^2_{x-2}(x)}{log_{x-2}(x)}\cdot \frac{1+log^2_{x-2}(x))}{log_{x-2}(x)}\leq 0 \end{matrix}\right.[/m]
При x >2; x ≠ 3
[m]1+log^2_{x-2}x >0[/m]
[m]log^2_{x-2}x >0[/m]
поэтому неравенство сводится к неравенству:
[m]1-log^2_{x-2}x ≤ 0 [/m]
[m]log^2_{x-2}x -1 ≥ 0 [/m]
[m](log_{x-2}x-1)( log_{x-2}x+1) ≥ 0 [/m]
__+___ [1-sqrt(2)] ____ [1+sqrt(2)] __+_
C учетом x >2; x ≠ 3 получаем ответ:
[1+sqrt(2);3)U(3;+ ∞ )