Answer:
Step-by-step explanation:
restrictions x≠±3
[tex]\frac{x-2}{x+3} +\frac{10x}{x^2-9} \\=\frac{(x-2)(x-3)}{(x+3)(x-3)} +\frac{10x}{x^2-9} \\=\frac{x^2-5x+6+10x}{x^2-9} \\=\frac{x^2+5x+6}{x^2-9} \\=\frac{x^2+3x+2x+6}{x^2-3^2} \\=\frac{x(x+3)+2(x+3)}{(x+3)(x-3)} \\=\frac{(x+3)(x+2)}{(x+3)(x-3)} \\=\frac{x+2}{x-3}[/tex]
