0} \atop {x-2>0}} \right.\\\\\left \{ {{(x-2)(x-3)>0} \atop {x>2}} \right.\\\\\left \{ {{x\in(-\infty;2)\cup((3;+\infty)} \atop {x>2}} \right.\Rightarrow x\in(3;+\infty)\\\\log_{5}(x^{2}-5x+6)-log_{5}(x-2)=1\\\\log_{5}\frac{(x-2)(x-3)}{x-2}=1\\\\log_{5}(x-3)=1,x\neq2\\\\x-3=5\\\\x=8\\\\Otvet:\boxed{8}" alt="ODZ:\\\\\left \{ {{x^{2} -5x+6>0} \atop {x-2>0}} \right.\\\\\left \{ {{(x-2)(x-3)>0} \atop {x>2}} \right.\\\\\left \{ {{x\in(-\infty;2)\cup((3;+\infty)} \atop {x>2}} \right.\Rightarrow x\in(3;+\infty)\\\\log_{5}(x^{2}-5x+6)-log_{5}(x-2)=1\\\\log_{5}\frac{(x-2)(x-3)}{x-2}=1\\\\log_{5}(x-3)=1,x\neq2\\\\x-3=5\\\\x=8\\\\Otvet:\boxed{8}" align="absmiddle" class="latex-formula">