Question

x-3asupra luix+3-x+3asupra lui3-x+12xasupra luix la patrat-9

Answer 1

 [tex]\frac{x-3}{x+3}-\frac{x+3}{3-x}+\frac{12x}{x^2-9}\;\Leftrightarrow\;\frac{x-3}{x+3}+\frac{x+3}{x-3}+\frac{12x}{x^2-9}=\\ =\frac{(x-3)^2+(x+3)^2+12x}{(x-3)(x+3)}=\frac{x^2-6x+9+x^2+6x+9+12x}{(x-3)(x+3)}=\\ =\frac{2(x^2+6x+9)}{(x-3)(x+3)}=\frac{2(x+3)^{\not{2}}}{(x-3)(x\not+3)}=\frac{2(x+3)}{x-3}[/tex]