ОДЗ: 
0\\7-x\neq1\\2x+9>0 \end{cases}\\\begin{cases} x<7\\x\neq6\\x>-4.5 \end{cases}" alt="\begin{cases} 7-x>0\\7-x\neq1\\2x+9>0 \end{cases}\\\begin{cases} x<7\\x\neq6\\x>-4.5 \end{cases}" align="absmiddle" class="latex-formula">
![log_{7-x}(2x+9)\leq0\\log_{7-x}(2x+9)\leq log_{7-x}1\\(2x+9-1)(7-x-1)\leq0\\(2x+8)(6-x)\leq0\\x\in (-\infty;-4]\cup[6;+\infty)](https://tex.z-dn.net/?f=log_%7B7-x%7D%282x%2B9%29%5Cleq0%5C%5Clog_%7B7-x%7D%282x%2B9%29%5Cleq+log_%7B7-x%7D1%5C%5C%282x%2B9-1%29%287-x-1%29%5Cleq0%5C%5C%282x%2B8%29%286-x%29%5Cleq0%5C%5Cx%5Cin+%28-%5Cinfty%3B-4%5D%5Ccup%5B6%3B%2B%5Cinfty%29 "log_{7-x}(2x+9)\leq0\\log_{7-x}(2x+9)\leq log_{7-x}1\\(2x+9-1)(7-x-1)\leq0\\(2x+8)(6-x)\leq0\\x\in (-\infty;-4]\cup[6;+\infty)")
Включая ОДЗ: ![x\in (-4.5;-4]\cup(6;7)](https://tex.z-dn.net/?f=x%5Cin+%28-4.5%3B-4%5D%5Ccup%286%3B7%29 "x\in (-4.5;-4]\cup(6;7)")
Ответ:
Правило перехода:
(f(x)-g(x))(h(x)-1)" alt="log_{h(x)}(f(x))(f(x)-g(x))(h(x)-1)" align="absmiddle" class="latex-formula">