Question

Ответье пожалуйста..​

Answer 1

$\bigg(\dfrac{x^3-27}{x^2-9}-\dfrac{9x}{x+3}\bigg):\bigg(1-\dfrac{6}{x+3}\bigg)=$

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

$=\dfrac{x^2+3x+9-9x}{x+3}:\dfrac{x+3-6}{x+3}=\dfrac{x^2-6x+9}{x+3}\cdot\dfrac{x+3}{x-3}=\dfrac{(x-3)^2}{x-3}=x-3$

$\bigg(\dfrac{16+9x^2}{16-9x^2}-\dfrac{3x}{4-3x}\bigg)(3x+4)=\dfrac{16+9x^2-3x(4+3x)}{(4-3x)(4+3x)}\cdot(3x+4)=$

$=\dfrac{16+9x^2-12x-9x^2}{4-3x}=\dfrac{4(4-3x)}{4-3x}=4$