![x\cdot\left(\dfrac{2x^2-3x+1}{(x+1)^2}-1\right)\leqslant0\\ x\cdot\dfrac{x^2-5x}{(x+1)^2}\leqslant0\\ \dfrac{x^2(x-5)}{(x+1)^2}\leqslant0\\ x\in(-\infty,-1)\cup(-1,5]](https://tex.z-dn.net/?f=x%5Ccdot%5Cleft%28%5Cdfrac%7B2x%5E2-3x%2B1%7D%7B%28x%2B1%29%5E2%7D-1%5Cright%29%5Cleqslant0%5C%5C+x%5Ccdot%5Cdfrac%7Bx%5E2-5x%7D%7B%28x%2B1%29%5E2%7D%5Cleqslant0%5C%5C+%5Cdfrac%7Bx%5E2%28x-5%29%7D%7B%28x%2B1%29%5E2%7D%5Cleqslant0%5C%5C+x%5Cin%28-%5Cinfty%2C-1%29%5Ccup%28-1%2C5%5D "x\cdot\left(\dfrac{2x^2-3x+1}{(x+1)^2}-1\right)\leqslant0\\ x\cdot\dfrac{x^2-5x}{(x+1)^2}\leqslant0\\ \dfrac{x^2(x-5)}{(x+1)^2}\leqslant0\\ x\in(-\infty,-1)\cup(-1,5]")
Найдем ОДЗ:
0\\x+1\ne1\\2x^2-3x+1>0\end{array}\right.\\ \left\lbrace\begin{array}{l}x+1>0\\x+1\ne1\\(2x-1)(x-1)>0\end{array}\right.\\ x\in(-1,0)\cup(0,\frac12)\cup(1,+\infty)" alt="\left\lbrace\begin{array}{l}x+1>0\\x+1\ne1\\2x^2-3x+1>0\end{array}\right.\\ \left\lbrace\begin{array}{l}x+1>0\\x+1\ne1\\(2x-1)(x-1)>0\end{array}\right.\\ x\in(-1,0)\cup(0,\frac12)\cup(1,+\infty)" align="absmiddle" class="latex-formula">
Пересекая оба множества, получим ответ
![x\in(-1,0)\cup(0,\frac12)\cup(1,5]](https://tex.z-dn.net/?f=x%5Cin%28-1%2C0%29%5Ccup%280%2C%5Cfrac12%29%5Ccup%281%2C5%5D "x\in(-1,0)\cup(0,\frac12)\cup(1,5]")