#9
x\\x\in (4;4+\sqrt{3})" alt="\tt\displaystyle (x-4)^2<\sqrt{3}(x-4)\\(x-4)^2-\sqrt{3}(x-4)<0\\(x-4)(x-4-\sqrt{3})<0\\+++(4)---(4+\sqrt{3})++++>x\\x\in (4;4+\sqrt{3})" align="absmiddle" class="latex-formula">
Ответ:
#10
x\\x\in [-1;1]" alt="(3x-7)^2\geq (7x-3)^2\\(3x-7)^2-(7x-3)^2\geq 0\\(3x-7-7x+3)(3x-7+7x-3)\geq 0\\(-4x-4)(10x-10)\geq 0\\(x+1)(x-1)\leq 0\\+++[-1]---[1]+++>x\\x\in [-1;1]" align="absmiddle" class="latex-formula">
Ответ:
#11
x\\x\in [-3;3]" alt="\tt\displaystyle x^2(-x^2-9)\leq 9(-x^2-9)\\x^2(x^2+9)\leq 9(x^2+9)\\x^2(x^2+9)-9(x^2+9)\leq 0\\(x^2-9)(x^2+9)\leq 0\\(x-3)(x+3)(x^2+9)\leq 0\\+++[-3]---[3]+++>x\\x\in [-3;3]" align="absmiddle" class="latex-formula">
Ответ:
#6
x\\x\in (-\infty;-1)\cup(7;+\infty)" alt="\sf\displaystyle \frac{-16}{x^2-6x-7} \leq 0|:(-16)\\\\ \frac{1}{x^2-6x-7}\geq 0\\ \frac{1}{(x-7)(x+1)}\geq 0\\x\neq 7; x\neq -1\\+++(-1)---(7)+++>x\\x\in (-\infty;-1)\cup(7;+\infty)" align="absmiddle" class="latex-formula">
Ответ:
#7
x\\x\in (4-\sqrt{6};4+\sqrt{6})" alt="\sf\displaystyle \frac{-13}{(x-4)^2-6}\geq 0|:(-13)\\\\ \frac{1}{(x-4)^2-6}\leq 0 \\\\ \frac{1}{(x-4-\sqrt{6})(x-4+\sqrt{6})}\leq 0\\x\neq 4+\sqrt{6};x\neq 4-\sqrt{6}\\+++(4-\sqrt{6})---(4+\sqrt{6})+++>x\\x\in (4-\sqrt{6};4+\sqrt{6})" align="absmiddle" class="latex-formula">
Ответ: