Question

СРОЧНО!!!!!! 20 БАЛЛОВ!!!!
$log_ \frac{x}{3} (3x - 2x + 1) \geqslant 0$
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Answer 1

$1)\quad log_{x/3}(3x-2x+1)\geq 0\; \; \to \; \; log_{x/3}(x+1)\geq 0\; ,\\\\ODZ:\; \left \{ {{x>-1} \atop {\frac{x}{3}>0\; ,\; \frac{x}{3}\ne 1}} \right.\; \; \left \{ {{x>-1} \atop {x>0\; ,\; x\ne 3}} \right.\; \; \to \; \; x\in (0,3)\cup (3,+\infty )\\\\(\frac{x}{3}-1)(x+1-1)\geq 0\\\\\frac{x\cdot (x-3)}{3}\geq 0\; \; \ ;\; \; +++[\, 0\, ]---[\, 3\, ]+++\\\\x\in (-\infty ,0\, ]\cup [\, 3,+\infty )\\\\\left \{ {{x\in (0,3)\cup (3,+\infty )} \atop {x\in (-\infty ,0\, ]\cup [\, 3,+\infty )}} \right. \; \; \; \to \; \; \; x\in (3,+\infty )$

$2)\quad log_{x/3}(3x^2-2x+1)\geq 0\\\\ODZ:\; \left \{ {{3x^2-2x+1>0} \atop {\frac{x}{3}>0\; ,\; \frac{x}{3}\ne 1}} \right.\; \; \left \{ {{x\in (-\infty ,+\infty )} \atop {x>0\; ,\; x\ne 3}} \right.\; \; \to \; \; x\in (0,3)\cup (3,+\infty )\\\\(\frac{x}{3}-1)(3x^2-2x+1-1)\geq 0\\\\\frac{x\cdot (3x-2)(x-3)}{3}\ge0\; \; \; \; \; ---[\, 0\, ]+++[\, \frac{2}{3}\, ]---[\, 3\, ]+++\\\\x\in [\, 0,\frac{2}{3}\, ]\cup [\, 3,+\infty )\\\\\left \{ {{x\in (0,3)\cup (3,+\infty )} \atop {x\in [\, 0,\frac{2}{3}\, ]\cup [\, 3,+\infty )}} \right.\\\\x\in (0,\frac{2}{3}\, ]\cup (3,+\infty )$