Question

Помогите пожалуста! Срочно нужен ответ!

Answer 1

$\dfrac4{1+x}-\dfrac{x+1}{x-1}=\dfrac{x^2-3}{1-x^2}\\\\\dfrac{4x-4-(1+x)(x+1)}{(x+1)(x-1)}-\dfrac{x^2-3}{1-x^2}=0\\\\\dfrac{4x-4-(x+1+x^2+x)}{x^2-1}-\dfrac{x^2-3}{1-x^2}=0\\\\\dfrac{4x-4-x-1-x^2-x}{x^2-1}-\dfrac{x^2-3}{-(x^2-1)}=0\\\\\dfrac{-x^2+2x-5}{x^2-1}+\dfrac{x^2-3}{x^2-1}=0\\\\\dfrac{-x^2+2x-5+x^2-3}{x^2-1}=0\qquad|\cdot(x^2-1)\\\\2x-8=0\\\\2x=8\\\\x=4$