Answer
$\dfrac{2+\dfrac{6}{x}}{1-\dfrac{9}{x^{2}}}=\dfrac{2x}{x-3}$
Work Step by Step
$\dfrac{2+\dfrac{6}{x}}{1-\dfrac{9}{x^{2}}}$
Evaluate the sum indicated in the numerator and the substraction indicated in the denominator:
$\dfrac{2+\dfrac{6}{x}}{1-\dfrac{9}{x^{2}}}=\dfrac{\dfrac{2x+6}{x}}{\dfrac{x^{2}-9}{x^{2}}}=...$
Evaluate the division:
$...=\dfrac{2x+6}{x}\div\dfrac{x^{2}-9}{x^{2}}=\dfrac{x^{2}(2x+6)}{x(x^{2}-9)}=...$
Take out common factor $2$ from the parentheses in the numerator and factor the parentheses in the denominator. After that, simplify:
$...=\dfrac{2x^{2}(x+3)}{x(x-3)(x+3)}=\dfrac{2x^{2}}{x(x-3)}=\dfrac{2x}{x-3}$
