0, a>0, \\ x\in(-\infty;0,4]\cup[2;+\infty)\cup\{0,5\}" alt="|4x^2-12x+5|(5x^2-12x+4) \geq 0, \\ |4x^2-12x+5| \geq 0\ \forall x\in R, \\ 4x^2-12x+5=0, \\ \frac{D}{4} =16, \\ x_1= \frac{1}{2}=0,5 , \ x_2= \frac{5}{2}=2,5; \\ 5x^2-12x+4 \geq 0, \\ 5x^2-12x+4 = 0, \\ \frac{D}{4} =16, \\ x_1= \frac{2}{5}=0,4 , \ x_2=2; \\ D>0, a>0, \\ x\in(-\infty;0,4]\cup[2;+\infty)\cup\{0,5\}" align="absmiddle" class="latex-formula">