ЕГЭ Профиль
Задание 2993
Решите неравенство $$\log_{\frac{5-x}{4}}(x-2)\cdot \log_{x-2}(6x-x^{2})\geq \log_{\frac{5-x}{4}}(3x^{2}-10x+15)$$
$$\left\{\begin{matrix}\frac{5-x}{4}>0\\\frac{5-x}{4}\neq1\\x-2\neq1\\x-2>0\\6x-x^{2}>0\\3x^{2}-10x+15>0\end{matrix}\right.$$ $$\Leftrightarrow$$ $$\left\{\begin{matrix}5-x>0\Rightarrow x<5\\x\neq1; x\neq3\\x>2\\x\in(0;6)\end{matrix}\right.$$ $$3x^{2}-10x+15>0$$ $$D=100-12\cdot15>0$$ $$x\in(2; 5)$$ $$\log_{\frac{5-x}{4}}(x-2)\cdot \log_{x-2}(6x-x^{2})\geq \log_{\frac{5-x}{4}}(3x^{2}-10x+15)$$ $$\frac{1}{\log_{x-2}\frac{5-x}{4}}\cdot \log_{x-2}(6x-x^{2})\geq \log_{\frac{5-x}{4}}(3x^{2}-10x+15)$$ $$\log_{\frac{5-x}{4}}(6x-x^{2})\geq \log_{\frac{5-x}{4}}(3x^{2}-10x+15)$$ $$(\frac{5-x}{4}-1)(6x-x^{2}-(3x^{2}-10x+15))\geq0$$ $$\frac{5-x-4}{4}\cdot(6x-x^{2}-3x^{2}+10x-15))\geq0$$ $$(1-x)\cdot(-4x^{2}+16x-15)\geq0$$ $$(x-1)\cdot(4x^{2}-16x+15)\geq0$$ $$D=256-240=16$$ $$x_{1}=\frac{16+4}{8}=2,5$$ $$x_{2}=\frac{16-4}{8}=1,5$$