ЕГЭ Профиль
Задание 2993
Решите неравенство $$\log_{\frac{5-x}{4}}(x-2)\cdot \log_{x-2}(6x-x^{2})\geq \log_{\frac{5-x}{4}}(3x^{2}-10x+15)$$
$$\left\{\begin{matrix}\frac{5-x}{4}>0\\\frac{5-x}{4}\neq1\\x-2\neq1\\x-2>0\\6x-x^{2}>0\\3x^{2}-10x+15>0\end{matrix}\right.$$ $$\Leftrightarrow$$ $$\left\{\begin{matrix}5-x>0\Rightarrow x<5\\x\neq1; x\neq3\\x>2\\x\in(0;6)\end{matrix}\right.$$ $$3x^{2}-10x+15>0$$ $$D=100-12\cdot15>0$$ $$x\in(2; 5)$$ $$\log_{\frac{5-x}{4}}(x-2)\cdot \log_{x-2}(6x-x^{2})\geq \log_{\frac{5-x}{4}}(3x^{2}-10x+15)$$ $$\frac{1}{\log_{x-2}\frac{5-x}{4}}\cdot \log_{x-2}(6x-x^{2})\geq \log_{\frac{5-x}{4}}(3x^{2}-10x+15)$$ $$\log_{\frac{5-x}{4}}(6x-x^{2})\geq \log_{\frac{5-x}{4}}(3x^{2}-10x+15)$$ $$(\frac{5-x}{4}-1)(6x-x^{2}-(3x^{2}-10x+15))\geq0$$ $$\frac{5-x-4}{4}\cdot(6x-x^{2}-3x^{2}+10x-15))\geq0$$ $$(1-x)\cdot(-4x^{2}+16x-15)\geq0$$ $$(x-1)\cdot(4x^{2}-16x+15)\geq0$$ $$D=256-240=16$$ $$x_{1}=\frac{16+4}{8}=2,5$$ $$x_{2}=\frac{16-4}{8}=1,5$$
Задание 3036
Решите неравенство $$\log_{3}(2^{x}+1)+\log_{2^{x}+1}3\geq 2,5$$
$$\log_{3}(2^{x}+1)+\log_{2^{x}+1}3\geq 2,5$$ $$\left\{\begin{matrix}2^{x}+1>0\\2^{x}+1\neq1\end{matrix}\right.$$ $$\left\{\begin{matrix}2^{x}>-1\\2^{x}\neq0\end{matrix}\right.$$ $$x\in R$$ $$\log_{3}(2^{x}+1)=y$$ $$y+\frac{1}{y}\geq\frac{5}{2}$$ $$\frac{y^{2}+1}{y}-\frac{5}{2}\geq0$$ $$\frac{2y^{2}+2-5y}{2y}\geq0$$ $$y\neq0$$ $$D=25-16=9$$ $$y_{1}=\frac{5+3}{4}=2$$ $$y_{2}=\frac{5-3}{4}=\frac{1}{2}$$ $$\left\{\begin{matrix}y>0\\y\leq\frac{1}{2}\\y\geq2\end{matrix}\right.$$ $$\left\{\begin{matrix}\log_{3}(2^{x}+1)>0\\\log_{3}(2^{x}+1)\leq\frac{1}{2}\\\log_{3}(2^{x}+1)\geq2\end{matrix}\right.$$ $$\Leftrightarrow$$ $$\left\{\begin{matrix}2^{x}+1>1\\2^{x}+1\leq\sqrt{3}\\2^{x}+1\geq9\end{matrix}\right.$$ $$\left\{\begin{matrix}x\in R\\x\leq\log_{2}(\sqrt{3}-1)\\x\geq3\end{matrix}\right.$$
Задание 4397
Решите неравенство: $$-3\log_{(x-1)}\frac{1}{3}+\log_{\frac{1}{3}}(x-1)>2|\log_{\frac{1}{3}}(x-1)|$$
ОДЗ: $$\left\{\begin{matrix}x-1>0\\x-1\neq1\end{matrix}\right.$$ $$\Leftrightarrow$$ $$x\in(1;2)\cup(2;+\infty)$$
$$\frac{-3}{\log_{\frac{1}{3}}(x-1)}+\log_{\frac{1}{3}}(x-1)-2|\log_{\frac{1}{3}}(x-1)|>0$$. Пусть $$\log_{\frac{1}{3}}(x-1)=y$$;
$$-\frac{3}{y}+y-2|y|>0$$ $$\left\{\begin{matrix}\left\{\begin{matrix}y\geq0\\-\frac{3}{y}-y>0\end{matrix}\right.(1)\\\left\{\begin{matrix}y<0\\-\frac{3}{y}+3y>0\end{matrix}\right.(2)\end{matrix}\right.$$ $$\Leftrightarrow$$
1) $$\left\{\begin{matrix}y\geq0\\\frac{-3-y^{2}}{y}>0\end{matrix}\right.$$ $$\Leftrightarrow$$ $$\left\{\begin{matrix}y\geq0\\-y^{2}>3\end{matrix}\right.$$ $$\Leftrightarrow$$ нет решений
2) $$\left\{\begin{matrix}y<0\\\frac{-1+y^{2}}{y}>0\end{matrix}\right.$$ $$\Leftrightarrow$$ $$\left\{\begin{matrix}y<0\\y^{2}<1\end{matrix}\right.$$ $$\Leftrightarrow$$ $$\left\{\begin{matrix}y<0\\y\in(-1;0)\end{matrix}\right.$$ $$\Leftrightarrow$$ $$\left\{\begin{matrix}\log_{\frac{1}{3}}(x-1)>-1\\\log_{\frac{1}{3}}(x-1)<0\end{matrix}\right.$$ $$\Leftrightarrow$$ $$\left\{\begin{matrix}x-1<3\\x-1>1\end{matrix}\right.$$ $$\Leftrightarrow$$ $$\left\{\begin{matrix}x<4\\x>2\end{matrix}\right.$$