[latex] frac{x+1}{x-3}+ frac{x-2}{x+1}= frac{x^2+x+12}{x^2-2x-3}\ Lewa strone sprowadzamy do wspolnego mianownika:\ frac{(x+1)(x+1)}{(x-3)(x+1)}+ frac{(x-2)(x-3)}{(x-3)(x+1)}= frac{x^2+x+12}{x^2-2x-3}\ frac{x^2+2 x+1}{(x-3)(x+1)}+ frac{x^2-5 x+6}{(x-3)(x+1)}= frac{x^2+x+12}{x^2-2x-3}\ frac{x^2+2x+1+x^2-5x+6}{(x-3)(x+1)}= frac{x^2+x+12}{x^2-2x-3}\ frac{2x^2-3x+7}{(x-3)(x+1)} = frac{x^2+x+12}{x^2-2x-3}\[/latex] [latex] frac{2x^2-3x+7}{(x+1)(x-3)}-frac{x^2+x+12}{x^2-2x-3^*}=0\ frac{2x^2-3x+7}{(x+1)(x-3)}- frac{x^2+x+12}{(x+1)(x-3)}=0\ frac{2x^2-3x+7-(x^2+x+12)}{(x+1)(x-3)}=0\ frac{2x^2-3x+7-x^2-x-12}{(x+1)(x-3)}=0\ frac{x^2-4x-5}{(x+1)(x-3)}=0\ Zalozenie:\ x+1 eq 0 oraz x-3 eq 0\ x eq -1 oraz x eq 3\ frac{x^2-4x-5^*}{(x+1)(x-3)}=0\ frac{(x+1)(x-5)}{(x+1)(x-3)}=0\ frac{x-5}{x-3}=0/*(x-3)\ (x-5)(x-3)=0\ x-5=0\ x=5\ [/latex] [latex]^* x^2-2x-3 mozna zapisc w postaci\ iloczynowej (x+1)(x-3)\\ ^* x^2-4x-5 mozna zapisac w postaci\ iloczynowej (x+1)(x-5)[/latex]
