A]
[tex]\frac{16-2x}{x^2-16} +\frac{3}{x+4} =\frac{16-2x}{x^2-16} +\frac{3}{x+4}\cdot\frac{x-4}{x-4} =\frac{16-2x}{x^2-16} +\frac{3(x-4)}{x^2-16}=\frac{16-2x+3x-12}{x^2-16} =\\\\\frac{4+x}{(x+4)(x-4)}=\frac{1}{x-4}[/tex]
założenie, że x jest różne od -4 i 4
B]
[tex]\frac{3x-2}{2x+4}+\frac{x+6}{x+2} =\frac{3x-2}{2x+4}+\frac{x+6}{x+2}\cdot\frac{2}{2}=\frac{3x-2}{2x+4}+\frac{2x+12}{2x+4} =\frac{3x-2+2x+12}{2x+4}=\\ \\\frac{5x+10}{2x+4}=\frac{5(x+2)}{2(x+2)}=\frac{5}{2}[/tex]
