[latex]Zad.1.\ a) frac{1}{x-1}=frac{2}{x+1}\\ x+1=2(x-1)\\ x+1=2x-2\\ x=3\\ b) frac{3}{x-2}=frac{2}{x-3}\\ 3(x-3)=2(x-2)\\ 3x-9=2x-4\\ x=5\\ c) frac{x+1}{x-1}=frac{x-5}{x-3}\\ (x+1)(x-3)=(x-1)(x-5)\\ x^2-3x+x-3=x^2-5x-x+5\\ -2x-3=-6x+5\\ 4x=8\\ x=2[/latex] [latex]Zad.2.\ a)\ x-4 eq0\ x eq4\\ 5+frac{1}{x-4}=frac{5-x}{x-4}\\ frac{5(x-4)+1}{x-4}=frac{5-x}{x-4}\\ 5(x-4)+1=5-x\\ 5x-20+1=5-x\\ 6x=24\\ x=4 sprzecznosc[/latex] [latex]b)\ x-5 eq0\ x eq5\\ frac{1}{x-5}+6=frac{6-x}{x-5}\\ frac{1+6(x-5)}{x-5}=frac{6-x}{x-5}\\ 1+6(x-5)=6-x\\ 1+6x-30=6-x\\ 7x=35\\ x=5 sprzecznosc[/latex] [latex]c)\ x-7 eq0\ x eq7\\ frac{8-x}{x-7}=8+frac{1}{x-7}\\ frac{8-x}{x-7}=frac{8(x-7)+1}{x-7}\\ 8-x=8x-56+1\\ 9x=63\\ x=7 sprzecznosc[/latex]
1a] x≠ 1∧x≠ -1 = x+1=2(x-1) x+1=2x-2 1+2=2x-x x=3 b] x≠2 ∧ x≠3 = 2(x-2)=3(x-3) 2x-4=3x-9 -4+9=3x-2x x=5 c] x≠1∧x≠3 = (x+1)(x-3)=(x-1)(x-5) x²-3x+x-3=x²-5x-x+5 -2x+6x=5+3 4x=8 x=2 2a] x≠4 5+1/(x-4)=(5-x)/(x-4) /*(x-4) 5(x-4)+1=1 5(x-4)=1-1 5(x-4)=0 5≠ 0 czyli x-4≠0, wiec x=4, ale xnie nalezy do dziedziny, wiec brak rozwiazania b] 1/(x-5)+6=(6-x)/(x-5) x≠5 1+6(x-5)=6-x 1+6x-30=6-x 6x+x=6+29 7x=35 x=35/7 x=5, ale 5 nie nalezy do D, czyli brak rozw. c] (8-x)/(x-7)=8+1/(x-7) x≠7 8-x=8(x-7)+1 8-x=8x-56+1 8+55=8x+x 9x=63 x=63/9 x=7, ale 7 nie nalezy do D, czyli brak rozw.
