\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{-16}{1-x^2}\left(x\ne\pm1\right)\)
\(\Leftrightarrow\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{16}{1-x^2}=0\)
\(\Leftrightarrow\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{16}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{16}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{x^2+2x+1-x^2+2x-1-16}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{4x-16}{\left(x-1\right)\left(x+1\right)}=0\)
=> 4x-16=0
<=> 4x=16
<=> x=4 (tmđk)
Vậy x=4
