[tex]\displaystyle \frac{x^2+3x}{x^2-9}=\frac{x(x+3)}{(x-3)(x+3)}=\frac{x}{x-3}[/tex]
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[tex]\displaystyle \frac{x^2+3x}{x^2-9}=\frac{x(x+3)}{(x-3)(x+3)}=\frac{x}{x-3}[/tex]