Ответ:
0\ \ ,\ \ \ \ x_1=2\ ,\ x_2=3\ \ (Viet)\\\\(x-2)(x-3)>0\ \ \ +++(2)---(3)+++\\\\x\in (-\infty ;\, 2\, )\cup (\, 3\, ;+\infty )" alt="log_{1/5}(x^2-5x+7)0\ ,\ \ x\in R\\\\a=\frac{1}{5}1\\\\x^2-5x+6>0\ \ ,\ \ \ \ x_1=2\ ,\ x_2=3\ \ (Viet)\\\\(x-2)(x-3)>0\ \ \ +++(2)---(3)+++\\\\x\in (-\infty ;\, 2\, )\cup (\, 3\, ;+\infty )" align="absmiddle" class="latex-formula">