100 БАЛЛОВ!!! Решите неравенство с логарифмами! Желательно с подробным решением!...

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$log_3(x^2-x-3)+log_3(2x^2+x-3) \geq log_3(x^2-2)^2+2+log_{\frac{1}{3}}4\\\\ODZ:\; \left \{ {{x^2-x-3\ \textgreater \ 0} \atop {2x^2+x-3\ \textgreater \ 0}} \right. \; \left \{ {{x\in (-\infty ,\frac{1-\sqrt{13}}{2})\cup (\frac{1+\sqrt{13}}{2},+\infty )} \atop {x\in (-\infty ,-\frac{3}{2})\cup (1,+\infty )}} \right. \to \\\\x\in (-\infty ,-\frac{3}{2})\cup (\frac{1+\sqrt{13}}{2},+\infty )$

$2=log_33^2=log_39; log_{\frac{1}{3}}4=-log_34\; ,\; \; \to \\\\log_3\left (((x^2-x-3)(2x^2+x-3)\right )-log_3\frac{9(x^2-2)^2}{4} \geq 0$

Mетод рационализации:

$(3-1)\cdot ((x^2-x-3)(2x^2+x-3)-\frac{9\cdot (x^2-2)^2}{4}) \geq 0\\\\2x^4-x^3-10x^2+9-\frac{9}{4}x^4+9x^2-9 \geq 0\\\\-\frac{1}{4}x^4-x^3-x^2 \geq 0\\\\-x^2(\frac{1}{4}x^2+x+1) \geq 0\\\\x^2(x^2+4x+4) \leq 0\\\\x^2(x+2)^2 \leq 0\; \; \to \; \; x=0\notin ODZ,\; \; x=-2\in ODZ\\\\Otvet:\; x=-2.\\\\--$

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