Question

$frac{16}{ x^{2} -16} + frac{x}{x+4} = frac{2}{x-4}$

Answer 1

$x neq 4 \ x neq -4 \ \ frac{16+x(x-4)-2(x+4)}{ x^{2} -16} =0 \ \ 16+ x^{2} -4x-2x-8=0 \ \ x^{2} -6x+8=0 \ \ (x-4)(x-2)=0 \ \ x=4;x=2$

x = 4 посторонний корень, не удовлетворяет ОДЗ

ОТВЕТ:
x = 2 

Answer 2

$frac{16}{ x^{2} -16} + frac{x}{x+4} =frac{2}{x-4} \ frac{16}{ (x-4)(x+4)} + frac{x}{x+4} =frac{2}{x-4} /* (x-4)(x+4) neq 0 \ (x-4)(x+4) neq 0 \ x-4 neq 0 \ x neq 4 \ x+4 neq 0 \ x neq -4 \ frac{16 (x-4)(x+4)}{ (x-4)(x+4)} + frac{x (x-4)(x+4)}{(x+4)} = frac{2 (x-4)(x+4)}{x-4} \ 16+x(x-4)=2(x+4)\16+ x^{2} -4x=2x+8 \ 16+ x^{2} -4x-2x-8=0$

$x^{2} -6x+8=0 \ D=36-32=4 \ sqrt{D} =2 \ x_{1} = frac{6+2}{2} = frac{8}{2} =4 \ x_{2} = frac{6-2}{2} = frac{4}{2} =2 \ \ \ \ \ x=2$