Question

Решите пж.$frac{1}{frac{1}{x + 4} + frac{1}{x} } = 126$

Answer 1

ОДЗ
x ≠ 0
x ≠ -4
x ≠ -2 

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