2.
0\\\\ 2t^2-19t+9=0\\\\ D=361-72=289=17^2\\\\ t_1=\frac{19-17}{4} =0,5\\\\ t_2=\frac{19+17}{4} =9\\\\ (5x-6)^2=0,5\\ 5x-6=-\frac{\sqrt{2}}{2}\\ x_1=\frac{12-\sqrt{2}}{10} \\\\ 5x-6=\frac{\sqrt{2}}{2}\\ x_2=\frac{12+\sqrt{2}}{10} \\\\ (5x-6)^2=9\\ 5x-6=-3\\ x_3=0,6\\\\ 5x-6=3\\ x_4=1,8 " alt=" (5x -6)^2=t>0\\\\ 2t^2-19t+9=0\\\\ D=361-72=289=17^2\\\\ t_1=\frac{19-17}{4} =0,5\\\\ t_2=\frac{19+17}{4} =9\\\\ (5x-6)^2=0,5\\ 5x-6=-\frac{\sqrt{2}}{2}\\ x_1=\frac{12-\sqrt{2}}{10} \\\\ 5x-6=\frac{\sqrt{2}}{2}\\ x_2=\frac{12+\sqrt{2}}{10} \\\\ (5x-6)^2=9\\ 5x-6=-3\\ x_3=0,6\\\\ 5x-6=3\\ x_4=1,8 " align="absmiddle" class="latex-formula">
3.
![\frac{x-3}{x(x+1)} \geq 0\\\\ -----(-1)---+---(0)-------[3]---+---\\\\ x \in(-1;0)U[3; +\infty)](https://tex.z-dn.net/?f=+%5Cfrac%7Bx-3%7D%7Bx%28x%2B1%29%7D+%5Cgeq+0%5C%5C%5C%5C+-----%28-1%29---%2B---%280%29-------%5B3%5D---%2B---%5C%5C%5C%5C+x+%5Cin%28-1%3B0%29U%5B3%3B+%2B%5Cinfty%29+ "\frac{x-3}{x(x+1)} \geq 0\\\\ -----(-1)---+---(0)-------[3]---+---\\\\ x \in(-1;0)U[3; +\infty)")