Question

resolva a equação quadrática (2 )/(x² - 4) - 1/(x - 2) + (x-4)/(x.(x+2)) = 0

Answer 1

Oi :)

$\frac{2}{x^2-4} - \frac{1}{x-2} + \frac{x-4}{x(x+2)} \\ \\ \frac{2}{x^2-2^2} - \frac{1}{x-2} + \frac{x-4}{x(x+2)} \\ \\ \frac{2}{(x+2)(x-2)} - \frac{1}{x-2} + \frac{x-4}{x(x+2)} \ \ \ agora \ tira \ mmc \\ \\ \frac{2.x-1.x(x+2)+(x-4)(x-2)}{x(x+2)(x-2)} \\ \\ \frac{2x-x^2-2x+x^2-2x-4x+8}{x(x+2)(x-2)} \\ \\ \frac{-6x+8}{x(x+2)(x-2)}$

Se for pra resolver o valor de x temos que igualar a zero:

$\frac{-6x+8}{x(x+2)(x-2)}=0 \ \ \ \ \ \ Multiplica \ ambos \ os \ lados \ por \ x(x+2)(x-2) \\ \\ \frac{-6x+8}{x(x+2)(x-2)}.x(x+2)(x-2)=0.x(x+2)(x-2) \\ \\ -6x+8=0 \\ \\ -6x=-8 \ \ *(-1) \\ \\ 6x=8 \\ x= \frac{8}{6} \\ \\ \boxed{x= \frac{4}{3}}$