w załącznikach rozwiązanie
[latex]a) left|frac{2x-2}{x-4} ight| > 4; D:x-4 eq0 o x eq4 o xinmathbb{R} ackslash {4}\\frac{2x-2}{x-4}=frac{2x-2-6+6}{x-4}=frac{2x-8+6}{x-4}=frac{2(x-4)+6}{x-4}=frac{6}{x-4}+frac{2(x-4)}{x-4}=frac{6}{x-4}+2\\y=frac{6}{x}\\Downarrow T_{vec{a}=[4;2]}\\y=frac{6}{x-4}+2\\Downarrow S_{OX dla y < 0}\\f(x)=left|frac{6}{x-4}+2 ight|\\Odp:xin(3; 4) cup (4; 7)[/latex] [latex]b) left|frac{x}{x+4} ight|leq3; D:x+4 eq0 o x eq-4 o xinmathbb{R} ackslash {-4}\\frac{x}{x+4}=frac{x+4-4}{x+4}=frac{-4}{x+4}+frac{x+4}{x+4}=frac{-4}{x+4}+1\\y=frac{-4}{x+4}\\Downarrow T_{vec{b}=[-4;1]}\\y=frac{-4}{x+4}+1\\Downarrow S_{OX dla y < 0}\\f(x)left|frac{-4}{x+4}+1 ight|\\Odp:xin(-infty;-6 > cup < -3; infty)[/latex] [latex]c) left|frac{x-1}{x-2} ight| > x-1; D:x-2 eq0 o x eq2 o xinmathbb{R} ackslash {2}\\frac{x-1}{x-2}=frac{x-1-1+1}{x-2}=frac{x-2+1}{x-2}=frac{1}{x+2}+frac{x-2}{x-2}=frac{1}{x+2}+1\\y=frac{1}{x}\\Downarrow T_{vec{c}=[-2;1]}\\y=frac{1}{x+2}+1\\Downarrow S_{OX dla y < 0}\\f(x)=left|frac{1}{x-2}+1 ight|\\Odp:xin(-infty; 1) cup (1; 2) cup (2; 3)\\\©DRK[/latex]
