0\\x=0\ ;x=2\\\_\_\_+\_\_\_(0)\_\_\_-\_\_\_(2)\_\_+\_\_>x\\x\in(-\infty;0)\cup(2;\infty)\\log_3(x^2-2x)>1\\x^2-2x>3\\x^2-2x-3=0\\x_1=3\ ;x_2=-1\\///+///(0)\_\_\_-\_\_\_(2)////+/////>x\\/+/(-1)\_\_\_\_\_\_\_-\_\_\_\_\_\_\_\_(3)//+//>x\\x\in(-\infty;-1)\cup(3;+\infty)" alt="10)OD3:\\x^2-2x>0\\x=0\ ;x=2\\\_\_\_+\_\_\_(0)\_\_\_-\_\_\_(2)\_\_+\_\_>x\\x\in(-\infty;0)\cup(2;\infty)\\log_3(x^2-2x)>1\\x^2-2x>3\\x^2-2x-3=0\\x_1=3\ ;x_2=-1\\///+///(0)\_\_\_-\_\_\_(2)////+/////>x\\/+/(-1)\_\_\_\_\_\_\_-\_\_\_\_\_\_\_\_(3)//+//>x\\x\in(-\infty;-1)\cup(3;+\infty)" align="absmiddle" class="latex-formula">
0\\6-x=0\ ;x+1=0\\x=6;x=-1\\\_\_-\_\_\_(-1)///+///(6)\_\_\_-\_\_>x\\log_\frac{1}{2}\frac{6-x}{x+1}\leq-2;\frac{1}{2}<1\\\frac{6-x}{x+1}\geq4\\\frac{6-x}{x+1}-4\geq0\\\frac{6-x-4x-4}{x+1}\geq0\\\frac{2-5x}{x+1}\geq0\\2-5x=0\ ;x+1=0\\x=0,4\ x=-1\\\_\_-\_\_\_(-1)///+///(6)\_\_\_-\_\_>x\\\_\_-\_\_\_(-1)/+/[0,4]\_\_\_-\_\_>x\\x\in(-1;0,4]" alt="11)OD3:\\\frac{6-x}{x+1}>0\\6-x=0\ ;x+1=0\\x=6;x=-1\\\_\_-\_\_\_(-1)///+///(6)\_\_\_-\_\_>x\\log_\frac{1}{2}\frac{6-x}{x+1}\leq-2;\frac{1}{2}<1\\\frac{6-x}{x+1}\geq4\\\frac{6-x}{x+1}-4\geq0\\\frac{6-x-4x-4}{x+1}\geq0\\\frac{2-5x}{x+1}\geq0\\2-5x=0\ ;x+1=0\\x=0,4\ x=-1\\\_\_-\_\_\_(-1)///+///(6)\_\_\_-\_\_>x\\\_\_-\_\_\_(-1)/+/[0,4]\_\_\_-\_\_>x\\x\in(-1;0,4]" align="absmiddle" class="latex-formula">
0\\3^x>8\\x>log_38\\log_3(3^x-8)=2-x\\3^x-8=3^{2-x}\\3^x-8=\frac{9}{3^x}|3^x\\3^{2x}-8*3^x-9=0\\3^x=9\ 3^x=-1\\x=2" alt="12)OD3:\\3^x-8>0\\3^x>8\\x>log_38\\log_3(3^x-8)=2-x\\3^x-8=3^{2-x}\\3^x-8=\frac{9}{3^x}|3^x\\3^{2x}-8*3^x-9=0\\3^x=9\ 3^x=-1\\x=2" align="absmiddle" class="latex-formula">
3^x=-1 решения не имеет
