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Интервалы изображены в приложении
Второе уравнение решается абсолютно также, как и на втором приложении.
![\bf\displaystyle\frac{\stackrel{\textnormal{D=4, x1=4, x2=2}}{(x^{2}-6x+8)}(x-2)^{3}}{(x+4)^{5}(5-x)^{3}}\geq 0 \implies \frac{(x-4)(x-2)^{4}}{-(x+4)^{5}(x-5)^{3}}\geq 0 \impies\\\\\\\implies\frac{(x-4)(x-2)^{4}}{(x+4)^{5}(x-5)^{3}}\leq 0 \implies (x-4)(x-2)^{4}(x+4)^{5}(x-5)^{3}\leq 0 \implies\\\\\\x\in(-\infty; -4)\cup[4; 5)\cup[2]](https://tex.z-dn.net/?f=%5Cbf%5Cdisplaystyle%5Cfrac%7B%5Cstackrel%7B%5Ctextnormal%7BD%3D4%2C%20x1%3D4%2C%20x2%3D2%7D%7D%7B%28x%5E%7B2%7D-6x%2B8%29%7D%28x-2%29%5E%7B3%7D%7D%7B%28x%2B4%29%5E%7B5%7D%285-x%29%5E%7B3%7D%7D%5Cgeq%200%20%5Cimplies%20%5Cfrac%7B%28x-4%29%28x-2%29%5E%7B4%7D%7D%7B-%28x%2B4%29%5E%7B5%7D%28x-5%29%5E%7B3%7D%7D%5Cgeq%200%20%5Cimpies%5C%5C%5C%5C%5C%5C%5Cimplies%5Cfrac%7B%28x-4%29%28x-2%29%5E%7B4%7D%7D%7B%28x%2B4%29%5E%7B5%7D%28x-5%29%5E%7B3%7D%7D%5Cleq%200%20%5Cimplies%20%28x-4%29%28x-2%29%5E%7B4%7D%28x%2B4%29%5E%7B5%7D%28x-5%29%5E%7B3%7D%5Cleq%200%20%5Cimplies%5C%5C%5C%5C%5C%5Cx%5Cin%28-%5Cinfty%3B%20-4%29%5Ccup%5B4%3B%205%29%5Ccup%5B2%5D "\bf\displaystyle\frac{\stackrel{\textnormal{D=4, x1=4, x2=2}}{(x^{2}-6x+8)}(x-2)^{3}}{(x+4)^{5}(5-x)^{3}}\geq 0 \implies \frac{(x-4)(x-2)^{4}}{-(x+4)^{5}(x-5)^{3}}\geq 0 \impies\\\\\\\implies\frac{(x-4)(x-2)^{4}}{(x+4)^{5}(x-5)^{3}}\leq 0 \implies (x-4)(x-2)^{4}(x+4)^{5}(x-5)^{3}\leq 0 \implies\\\\\\x\in(-\infty; -4)\cup[4; 5)\cup[2]")
