0\\ \\ x(5x-4)>0\\ \\ ++++(0)-----(\frac{4}{5} )++++\\ \\ x\in(-oo;0)U(0.8;+oo)\\ \\ metod\ intervalow" alt="4x<5x^2\\ \\ 5x^2-4x>0\\ \\ x(5x-4)>0\\ \\ ++++(0)-----(\frac{4}{5} )++++\\ \\ x\in(-oo;0)U(0.8;+oo)\\ \\ metod\ intervalow" align="absmiddle" class="latex-formula">