Verified answer

[tex] \frac{3}{ {x}^{2} + 8x + 16} = 1 + \frac{2}{x + 4} \\ 1 + \frac{2}{x + 4} - \frac{3}{ {(x + 4)}^{2} } - = 0 \\ \frac{ {x}^{2} + 8x + 16 + 2(x + 4) - 3}{{(x + 4)}^{2} } = 0 \\ \frac{ {x}^{2} + 10x + 21}{(x + 4) ^{2} } = 0 \\ \frac{(x + 3)(x + 7)}{ {(x + 4)}^{2} } = 0 \\ x_1 = - 3 \\ x _2 = - 7[/tex]

ответ:

[tex]x_1 = - 3 \\ x _2 = - 7[/tex]