ОДЗ :
x² - 5x + 8 > 0
x² - 5x + 8 = 0
D = (- 5)² - 4 * 8 = 25 - 32 = - 7 < 0
Дискриминант меньше нуля, старший коэффициент равен 1 > 0 , значит
x² - 5x + 8 > 0 при всех действительных значениях x .
![\frac{2}{5}^{log_{0,25}(x^{2}-5x+8)}\leq 2,5\\\\\frac{2}{5}^{log_{0,25}(x^{2}-5x+8)}\leq (\frac{2}{5})^{-1}\\\\0 \frac{2}{5}^{log_{0,25}(x^{2}-5x+8)}\leq 2,5\\\\\frac{2}{5}^{log_{0,25}(x^{2}-5x+8)}\leq (\frac{2}{5})^{-1}\\\\0](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B5%7D%5E%7Blog_%7B0%2C25%7D%28x%5E%7B2%7D-5x%2B8%29%7D%5Cleq%202%2C5%5C%5C%5C%5C%5Cfrac%7B2%7D%7B5%7D%5E%7Blog_%7B0%2C25%7D%28x%5E%7B2%7D-5x%2B8%29%7D%5Cleq%20%28%5Cfrac%7B2%7D%7B5%7D%29%5E%7B-1%7D%5C%5C%5C%5C0%3C%5Cfrac%7B2%7D%7B5%7D%20%3C1%5CRightarrow%20log_%7B%5Cfrac%7B1%7D%7B4%7D%7D%28x%5E%7B2%7D-5x%2B8%29%5Cgeq%20-1%5C%5C%5C%5C0%3C%5Cfrac%7B1%7D%7B4%7D%20%3C1%20%5CRightarrow%5C%5C%5C%5Cx%5E%7B2%7D-5x%2B8%5Cleq4%5C%5C%5C%5Cx%5E%7B2%7D-5x%2B4%5Cleq0%5C%5C%5C%5C%28x-1%29%28x-4%29%5Cleq%200)
+ - +
_____[1]______[4]______
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Ответ : x ∈ [1 ; 4]