Question

$( \frac{1}{x + 3} - \frac{27}{ {x}^{2} + 27 } + \frac{9}{ {x}^{2} - 3a + 9 }) \times (x - \frac{6x - 9}{x + 3} )$
8 класс
решите пж

Answer 1

$\displaystyle \tt {} \ \ \ \bigg(\frac{1}{x+3}- \frac{27}{x^{3}+27}+ \frac{9}{x^{2}-3x+9}\bigg)\cdot\bigg(x- \frac{6x-9}{x+3}\bigg)=\\\\\\=\frac{x^{2}-3x+9-27+9x+27}{(x+3)(x^{2}-3x+9)}\cdot \frac{x^{2}+3x-6x+9}{x+3}=\\\\\\=\frac{x^{2}+6x+9}{(x+3)(x^{2}-3x+9)}\cdot \frac{x^{2}-3x+9}{x+3}=\frac{(x+3)^{2}}{(x+3)^{2}}=1;$