Question

(9 разд.на x-3) +(18 разд.на x +3) -3 =0

Answer 1

$( \frac{9}{x-3} )+( \frac{18}{x+3} )-3 = 0$
$( \frac{9*(x+3)}{(x-3)(x+3)} )+ ( \frac{18*(x-3)}{(x-3)(x+3)} )- \frac{3(x-3)(x+3)}{(x-3)(x+3)} =0$
$\frac{9*(x+3)+18*(x-3)(-3)*(x-3)(x+3)}{(x-3)(x+3)} =0$
$\frac{9x+27+18x-54(-3)*(x^2-9)}{(x-3)(x+3)} =0$
$\frac{27x-27-3x^2+27}{(x-3)(x+3)} =0$
$\frac{-3x^2+27x}{(x-3)(x+3)} =0$      $x \neq 3$;$x \neq -3$
$-3x^2+27x=0$
$3x(-x+9)=0$
$3x=0$ или $-x+9=0$
$x_1=0$        $-x=-9$
                   $x_2=9$