Question

Помогите пожалуйста решить рациональные дроби
$3y - 6 \\ \frac{m - 4}{7 } \\ \frac{7}{m - 4 } \\ \frac{c - 8}{c + 10} \\ \frac{12}{x { }^{2} - 3c } \\ \frac{9}{x {}^{6} + 1 } \\ \frac{7}{ |x| - 8 } \\ \frac{x}{ |x| + 4 } \\ \frac{x - 1}{x {}^{2} + 10x + 25} \\ \frac{c }{c - 3} - \frac{c}{c + 4}$
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Answer 1

$3y-6\; \; ;\; \; y\in R\; \; ,\; \; R=(-\infty ,+\infty )\\\\\frac{m-4}{7}\; \; ;\; \; \; m\in R\\\\\frac{7}{m-4}\; \; ;\; \; \; m\ne 4\; \; ;\; \; m\in (-\infty ,4)\cup (4,+\infty )\\\\\frac{c-8}{c+10}\; \; ;\; \; \; c\ne -10\; \; ,\; \; c\in (-\infty ,-10)\cup (-10,+\infty )\\\\\frac{12}{x^2-3c}\; \; ;\; \; x^2\ne 3c\; ,\; \; x\ne \pm \sqrt{3c}\; ,\\\\x\in (-\infty ,-\sqrt{3c})\cup (-\sqrt{3c}\, ,\sqrt{3c})\cup (\sqrt{3c},+\infty )\; ,\; c\geq 0\\\\\frac{9}{x^6+1}\; \; ;\; \; x\in R\\\\\frac{7}{|x|-8}\; \; ;\; \; |x|\ne 8\; ,\; \; x\ne \pm 8\; ,\; x\in (-\infty ,-8)\cup (-8,8)\cup (8,+\infty )\\\\\frac{x}{|x|+4}\; \; ;\; \; x\in R\\\\\frac{x-1}{x^2+10x+25}=\frac{x-1}{(x+5)^2}\; \; ;\; \; x\ne -5\; ,\; \; x\in (-\infty ,-5)\cup (-5,+\infty )\\\\\frac{c}{c-3}-\frac{c}{c+4}\; \; ;\; \; c\ne 3\; ,\; c\ne -4\; ,\; c\in (-\infty ,-4)\cup (-4,3)\cup (3,+\infty )$