Question

преобразуйте в алгебраическую дробь:
$\frac{1}{x-3} + \frac{1}{x+3} + \frac{6}{ x^{2} -9}$

Answer 1

$\frac{1}{x-3} + \frac{1}{x+3} + \frac{6}{x^2-9} = \frac{1}{x-3} + \frac{1}{x+3} + \frac{6}{(x-3)(x+3)} = \\ \\ = \frac{x+3+x-3+6}{(x-3)(x+3)} = \frac{2x+6}{(x-3)(x+3)}$

Answer 2

$\frac{1}{x-3}+ \frac{1}{x+3}+ \frac{6}{ x^{2}-9 }= \frac{x+3+x-3+6}{ x^{2}-9 }= \frac{2x+6}{ x^{2}-9 }= \frac{2*(x+3)}{(x+3)*(x-3)}= \frac{2}{x-3}$