ПОМОГИТЕ ПОЖАЛУЙСТА.
2/x-3 - x/x-3 = 4x/x^2 - 9

$frac{2}{x-3}$ - $frac{x}{x-3}$ = $frac{4x}{ x^{2} -9}$
$frac{2-x}{x-3}$ - $frac{4x}{ x^{2} -9}$ =0
$frac{(x+3)(2-x)-4x}{(x-3)(x+3)}$ = $frac{2x- x^{2} +6-3x-4x}{(x-3)(x+3)}$ = $frac{-5x- x^{2} +6}{(x-3)(x+3)}$ = $frac{- x^{2} -5x+6}{(x-3)(x+3)}$
По Th Виета: $left { {{ x_{1} * x_{2} =c=6} atop { x_{1}+ x_{2} =-b=5 }} right.$
$left { {{ x_{1} =2} atop { x_{2}=3}} right.$
$frac{-(x-3)(x-2)}{(x-3)(x+3)}$ =- $frac{x-2}{x+3}$