Помогите решить логарифмическое неравенство [tex]log_{6x^{2} +5x} (2x^{2} -3x+1)\geq 0[/tex]. Created by Wumpscut. algebra-ru

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Помогите решить логарифмическое неравенство

[tex]log_{6x^{2} +5x} (2x^{2} -3x+1)\geq 0[/tex]

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NNNLLL54

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Метод рационализации: $log_{g}f\vee 0\; \; \Leftrightarrow \; \; (g-1)(f-1)\vee 0$ .

$log_{6x^2+5x}(2x^2-3x+1)\geq 0\\\\ODZ:\; \; \left \{ {{2x^2-3x+1>0} \atop {6x^2+5x>0,\; 6x^2+5x\ne 1}} \right. \; \left \{ {{2(x-1)(x-0,5)>0} \atop {x(6x+5)>0,\; 6(x+1)(x-\frac{1}{6})\ne 0}} \right. \\\\\left \{ {{x\in (-\infty ;\, 0,5)\cup (1,+\infty )\qquad \qquad } \atop {x\in (-\infty ;-\frac{5}{6})\cup (0,+\infty )\; ,\; x\ne -1\; ,\; x\ne \frac{1}{6}}} \right. \\\\x\in (-\infty ,-1)\cup (-1,-\frac{5}{6})\cup (\frac{1}{6},\frac{1}{2})\cup (1,+\infty )\\\\\\(2x^2-3x+1-1)(6x^2+5x-1)\geq 0$

$x(2x-3)\cdot 6\cdot (x+1)(x-\frac{1}{6})\geq 0\\\\x_1=0\; ,\; x_2=1,5\; ,\; x_3=-1\; ,\; x_4=\frac{1}{6}\\\\+++(-1)---(0)+++(\frac{1}{6})---[\, 1,5\, ]+++\\\\x\in (-\infty ,-1)\cup (0,\frac{1}{6})\cup [\, 1,5\, ;\, +\infty )\\\\\left \{ {{x\in (-\infty ,-1)\cup (0,\frac{1}{6})\cup [\, 1,5\, ;\, +\infty )} \atop {x\in (-\infty ,-1)\cup (-1,-\frac{5}{6})\cup (\frac{1}{6},\frac{1}{2})\cup (1,+\infty )}} \right. \; \; \Rightarrow \; \; \underline {x\in (-\infty ,-1)\cup [\, 1,5\, ;\, +\infty )}$

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