Question

Решите задачи даю 49 баллов...

Answer 1

$1)\; \; -x^2+5x-16>0\; \; \to \; \; x^2-5x+16<0\\\\D=\underline {-39<0}\; \; \to \; \; x^2-5x+16>0\; \; \to \; \; otvet:\; x\in \varnothing \\\\2)\; \; x^2-4\geq 0\; \; \to \; \; (x-2)(x+2)\geq 0\\\\+++[-2\, ]---[\, 2\, ]+++\\\\x\in (-\infty ,-2\, ]\cup [\, 2,+\infty )\\\\3)\; \; (x+6)(x+3)(x-6)(x-9)\geq 0\\\\+++[-6]---[-3]+++[\, 6\, ]---[\, 9\, ]+++\\\\x\in (-\infty ,-6\, ]\cup [-3,6\, ]\cup [\, 9,+\infty )\\\\4)\; \; (x-2)^2(x^2-9)\leq 0\; \; \to \; \; (x-2)^2(x-3)(x+3)\leq 0\\\\+++[-3\, ]---[\, 2\, ]---[\, 3\, ]+++$

$x\in [-3,3\, ]\\\\5)\; \; \frac{4-x}{x-8}\leq 0\; \; \to \; \; \frac{x-4}{x-8}\geq 0\\\\+++[\, 4\, ]---[\, 8\, ]+++\\\\x\in (-\infty ,4\, ]\cup [\, 8,+\infty )\\\\6)\; \; y=\sqrt{\frac{2-x}{x+3}}\\\\OOF:\; \; \frac{2-x}{x+3}\geq  0\; \; \to \; \; \frac{x-2}{x+3}\leq 0\\\\+++[-3\, ]---[\, 2\, ]+++\\\\x\in [-3,2\, ]$