nks/30598-2d3b98638e1ff9de.js","35243","static/chunks/35243-51da85fc2282ee1a.js","5322","static/chunks/5322-d651b874c6d06af9.js","36008","static/chunks/36008-17b732f356368970.js","31168","static/chunks/31168-535392cb7f9217f4.js","76986","static/chunks/app/(unauthorized-user)/testing/%5BexamType%5D/%5B...slug%5D/page-da791188224a5b92.js"],""] 54:T4c0, $\circledast\begin{cases}\left(x+3\right)^2>0\\\left(x+3\right)^2\neq1\\2x^2+9x+21>0\\x^2-x>0\end{cases}\Leftrightarrow\begin{cases}x+3\neq0\\(x+3-1)(x+3+1)\neq0\\D<0\Rightarrow x\in\mathbb{R}\\x(x-1)>0\end{cases}\Leftrightarrow\begin{cases}x\neq-3\\(x+2)(x+4)\neq0\\x\in(-\infty;0)\cup(1;+\infty)\end{cases}\Leftrightarrow$ $\Leftrightarrow\begin{cases}x\neq-3\\x\neq-2\\x\neq-4\\x\in(-\infty;0)\cup(1;+\infty)\end{cases}\Leftrightarrow x\in(-\infty;-4)\cup(-4;-3)\cup(-3;-2)\cup(-2;0)\cup(1;+\infty)$

Применим метод рационализации: $\left(\left(x+3\right)^2-1\right)\left(2x^2+9x+21-x^2+x\right)\geq0\Leftrightarrow\left(x+2\right)\left(x+4\right)\left(x^2+10x+21\right)\geq0\Leftrightarrow (x+2)(x+4)(x+3)(x+7)\geq0$

Решим с помощью метода интервалов: $x\in(-\infty;-7]\cup[-4;-3]\cup[-2;+\infty)$

Пересекая со $\circledast$ , получим: $x\in(-\infty;-7]\cup(-4;-3)\cup(-2;0)\cup(1;+\infty)$

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Задание №4201 ЕГЭ по Математике