założenia: D; x∈R {-5,1} [latex]frac{x-1}{x+5}= frac{x-3}{x-1} \ (x-1)^2=(x-3)(x+5)\ x^2-2x+1=x^2+2x-15\ -4x=-16\ x=4[/latex] [latex]18x^{3}+ 9x^{2}-18x-9=0\ 18x(x^2-1) +9(x^2-1)=0\ (18x+9)(x^2+1)=0\ 9(2x+1)(x-1)(x+1)=0\ 2x+1=0 vee x-1=0 vee x+1=0\ x=-frac{1}{2} vee x=1 vee x=-1[/latex]
[latex] frac{x-1}{x+5}= frac{x-3}{x-1} D=mathbb{R} setminus {-5;1}\ (x+5)(x-3)=(x-1)(x-1)\ x^2-3x+5x-15=x^2-x-x+1\ x^2+2x-15=x^2-2x+1\ 4x-16=0\ 4x=16\ x=4\[/latex] [latex]18x^3+9x^2-18x-9=0/:9\ 2x^3+x^2-2x-1=0\ x^2(2x+1)-(2x+1)=0\ (2x+1)(x^2-1)=0\ (2x+1)(x-1)(x+1)=0\ 2x+1=0 x=1 x=-1\ x=- frac{1}{2} x=1 x=-1\ [/latex]
