Question

Answer 1

đkxđ : x khác 1, x khác -1

<=> \(\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}-\dfrac{4x^2+1}{\left(x-1\right)\left(x+1\right)}=0\)

<=> \(\dfrac{x^2+2x+1}{x^2-1}-\dfrac{x^2-2x+1}{x^2-1}-\dfrac{4x^2+1}{x^2-1}=0\)

<=> \(x^2+2x+1-x^2+2x-1-4x^2-1=0\)

<=> \(-4x^2+4x-1=0\)

<=> Áp dụng ct bậc 2:

\(x=\dfrac{-\left(-4\right)\pm\sqrt{\left(-4\right)^2-4.4.1}}{2.4}\)

= \(\dfrac{1}{2}\left(nhận\right)\)

Vậy..