Question

рациональные дроби,помогите решить
3y-6
m-4/7
7/m-4
c-8/c+10
12/x²-3c
9/x^6+1
7/|x|-8
x/|x|+4
x-1/x²+10x+25
c/c-3 - c/c+4​

Answer 1

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