Question

(x^2+27/6x^2-54 + 1/3-x) : x-3/6

Answer 1

$(frac{x^2+27}{6x^2-54}+frac{1}{3-x}):frac{x-3}{6}= \ \=(frac{x^2+27}{6(x^2-9)}+frac{1}{-x+3}):frac{x-3}{6}= \ \=(frac{x^2+27}{6(x-3)(x+3)}-frac{1}{1-3}):frac{x-3}{6}= \ \=frac{x^2+27-6(x+3)}{6(x+3)(x-3)}*frac{6}{x-3}= \ \=frac{x^2+27-6x-18}{6(x+3)(x-3)}*frac{6}{x-3}= \ \=frac{x^2-6x+9}{6(x+3)(x-3)}*frac{6}{x-3}= \ \=frac{(x-3)(x-3)}{6(x+3)(x-3)}*frac{6}{x-3}= \ \=frac{6(x-3)}{6(x-3)(x+3)}= \ \=frac{1}{x+3}$

 

ну как то вот так....