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$\circledast \begin{cases}x\ne-2\\x\ne1\end{cases}$

$\frac{x^4-2x^3+x^2}{x^2+x-2}-\frac{2x^3+x^2+x-1}{x+2}\leq1$ $\frac{x^{2}(x^{2}-2x+1)}{(x+2)(x-1)}-\left(\frac{2x^{3}+x^{2}+x-1}{x+2}+\frac{x+2}{x+2}\right)\leq0$ $\frac{x^{2}(x-1)^2}{(x+2)(x-1)}-\frac{2x^{3}+x^{2}+2x+1}{x+2}\leq0$ $\frac{x^{2}(x-1)}{x+2}-\frac{x^{2}(2x+1)+(2x+1)}{x+2}\leq0$ $\frac{x^{2}(x-1)}{x+2}-\frac{(2x+1)(x^{2}+1)}{x+2}\leq0$ $\frac{x^3-x^2-(2x^3+x^2+2x+1)}{x+2}\leq0$ $\frac{x^{3}-x^2-2 x^{3}-x^2-2x-1}{x+2}\leq0$ $\frac{-x^{3}-2x^{2}-2x-1}{x+2}\leq0\mid\cdot(-1)$ $\frac{x^3+2x^2+2x+1}{x+2}\geq0$ $\frac{(x+1)(x^{2}+x+1)}{x+1}\geq0\Leftrightarrow \left[\begin{matrix}x<-2\\-1\le x<1\\x>1\end{matrix}\right.x\in(-\infin;-2)\cup[-1;1)\cup(1;+\infin)$ 46:{"original_url":"https://storage.yandexcloud.net/egehub/public/776805/xROJAaiT4TVwNm1CfpuKJtAEq4ygpFHyqpoUsCEs.png","small_url":"https://storage.yandexcloud.net/egehub/public/776805/conversions/xROJAaiT4TVwNm1CfpuKJtAEq4ygpFHyqpoUsCEs-small.png","medium_url":"https://storage.yandexcloud.net/egehub/public/776805/conversions/xROJAaiT4TVwNm1CfpuKJtAEq4ygpFHyqpoUsCEs-medium.png","big_url":"https://storage.yandexcloud.net/egehub/public/7

Задание №4188 ЕГЭ по Математике