Ответ:
Объяснение:
0 ;x<7\\7-x\leq 0,6^{2} \\7-x\leq 0,36\\x\geq 6,64\\" alt="A)\\logx_{0,6} (7-x)\geq 2 ;\\ODZ:7-x>0 ;x<7\\7-x\leq 0,6^{2} \\7-x\leq 0,36\\x\geq 6,64\\" align="absmiddle" class="latex-formula">
Ответ: x∈[6,64;7).
Б)
0; x>-0,86\\x+0,86\leq 7^{1} \\x+0,86\leq 7\\x\leq 6,14." alt="log_{7} (x+0,86)\leq 1\\ODZ:x+0,86>0; x>-0,86\\x+0,86\leq 7^{1} \\x+0,86\leq 7\\x\leq 6,14." align="absmiddle" class="latex-formula">
Ответ: х∈(-0,86;6,14]
B)

Г)

-∞__+__-1__-__3,5__+__+∞
x∈(-1;3,5).
Д)
0;\\(x-8)(6-x)\geq 0\\" alt="\frac{(x-8)(6-x)}{(2-x)^{2} }\geq 0\\ODZ:(2-x)^{2} \neq 0;2-x\neq 0;x\neq 2\\(2-x)^{2} >0;\\(x-8)(6-x)\geq 0\\" align="absmiddle" class="latex-formula">
-∞_-_(2)_-_6__+__8__-__+∞
x∈[6;8].