Question

Упростить выражение
x^2+3x/(x-3)^2 : ( 3/x+3 + x^2+9/x^2-9 - 3/3-x )

Answer 1

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