Question

 \(\dfrac{x+1}{x-1}\)-\(\dfrac{x-1}{x+1}\)=\(\dfrac{16}{x^2-1}\)

 

Answer 1

222222222+2x−1=16

=>4x=16=>x=4

Chúc bạn học tốt

Answer 2

Answer 3

ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2-1}\)

\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{16}{\left(x-1\right)\left(x+1\right)}\)

Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=16\)

\(\Leftrightarrow x^2+2x+1-x^2+2x-1-16=0\)

\(\Leftrightarrow4x-16=0\)

\(\Leftrightarrow4x=16\)

hay x=4(thỏa ĐK)

Vậy: S={4}