Решите неравенство . решите, пожалуйста, подробно
https://znanija.com/task/27278931
пусть х=0: 2*|-4|/(-1) разве не < 0 ? х=0 тоже решение...
![\displaystyle\mathtt{\frac{(x^2-3x+2)|x-4|}{x^2-1}\leq0;~\left\{{{\frac{(x^2-3x+2)(x-4)}{x^2-1}\leq0,}\atop{x\geq4;}}\right\left\{{{x\in(-1;1)U[2;4],}\atop{x\geq4;}}\right}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathtt%7B%5Cfrac%7B%28x%5E2-3x%2B2%29%7Cx-4%7C%7D%7Bx%5E2-1%7D%5Cleq0%3B%7E%5Cleft%5C%7B%7B%7B%5Cfrac%7B%28x%5E2-3x%2B2%29%28x-4%29%7D%7Bx%5E2-1%7D%5Cleq0%2C%7D%5Catop%7Bx%5Cgeq4%3B%7D%7D%5Cright%5Cleft%5C%7B%7B%7Bx%5Cin%28-1%3B1%29U%5B2%3B4%5D%2C%7D%5Catop%7Bx%5Cgeq4%3B%7D%7D%5Cright%7D "\displaystyle\mathtt{\frac{(x^2-3x+2)|x-4|}{x^2-1}\leq0;~\left\{{{\frac{(x^2-3x+2)(x-4)}{x^2-1}\leq0,}\atop{x\geq4;}}\right\left\{{{x\in(-1;1)U[2;4],}\atop{x\geq4;}}\right}")
![\mathtt{x\in(-\infty;-1)U(1;2]U[4]}](https://tex.z-dn.net/?f=%5Cmathtt%7Bx%5Cin%28-%5Cinfty%3B-1%29U%281%3B2%5DU%5B4%5D%7D "\mathtt{x\in(-\infty;-1)U(1;2]U[4]}")
![\displaystyle\mathtt{\frac{(x^2-3x+2)|x-4|}{x^2-1}\leq0;~\left\{{{x\in(-\infty;-1)U(1;2]U[4;+\infty),}\atop{x\leq4;}}\right}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cmathtt%7B%5Cfrac%7B%28x%5E2-3x%2B2%29%7Cx-4%7C%7D%7Bx%5E2-1%7D%5Cleq0%3B%7E%5Cleft%5C%7B%7B%7Bx%5Cin%28-%5Cinfty%3B-1%29U%281%3B2%5DU%5B4%3B%2B%5Cinfty%29%2C%7D%5Catop%7Bx%5Cleq4%3B%7D%7D%5Cright%7D "\displaystyle\mathtt{\frac{(x^2-3x+2)|x-4|}{x^2-1}\leq0;~\left\{{{x\in(-\infty;-1)U(1;2]U[4;+\infty),}\atop{x\leq4;}}\right}")