120-ға көбейтеміз: \[12(9x+1)-3(x-2)>2(5x+1)-8(8x-2).\] \[108x+12-3x+6>10x+2-64x+16.\] \[105x+18>-54x+18\Rightarrow159x>0\Rightarrow x>0.\]
Жауабы: \((0;+\infty)\).