Решите пж.

ОДЗ
x ≠ 0
x ≠ -4
x ≠ -2

$\dfrac{1}{ \dfrac{1}{x+4} + \dfrac{1}{x} } = 126 \\ \\ \\ \dfrac{1}{ \dfrac{x + x + 4}{x^2 + 4x} } = 126 \\ \\ \\ \dfrac{x^2 + 4x}{2x + 4} = 126 \\ \\ x^2 + 4x = 126(2x + 4) \\ \\ x^2 + 4x = 252x + 504 \\ \\ x^2 - 248x - 504 = 0 \\ \\ x^2 - 248x + 15376 - 15880 = 0 \\ \\ (x - 124)^2 - (2 \sqrt{3970})^2 = 0 \\ \\ (x - 124 - 2 \sqrt{3970} )(x - 124 + 2 \sqrt{3970} ) = 0 \\ \\ x = 124 + 2 \sqrt{3970} \ \ \ \ \ \ \ and \ \ \ \ \ x = 124 - 2 \sqrt{3970}$