$a) \dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2-1}$
$⇔\dfrac{(x+1)^2}{(x-1)(x+1)}-\dfrac{(x-1)^2}{(x+1)(x-1)}=\dfrac{16}{(x+1)(x-1)}$
ĐKXĐ: `x ne ± 1`
`⇒(x+1)^2-(x-1)^2=16`
`⇔x^2+2x+1-(x^2-2x+1)=16`
`⇔x^2+2x+1-x^2+2x-1=16`
`⇔x^2+2x+1-x^2+2x-1-16=0`
`⇔4x-16=0`
`⇔4(x-4)=0`
`⇔x-4=0`
`⇔x=4(N)`
Vậy `S={4}`
$b)\dfrac{12}{x^3+8}=1+\dfrac{1}{x+2}$
ĐKXĐ: `x ne -2`
$⇔\dfrac{12}{(x+2)(x^2-2x+4)}=\dfrac{(x+2)(x^2-2x+4)}{(x+2)(x^2-2x+4)} +\dfrac{x^2-2x+4}{(x+2)(x^2-2x+4)}$
`⇒(x+2)(x^2-2x+4)+x^2-2x+4=12`
`⇔x^3+8+x^2-2x+4=12`
`⇔x^3+8+x^2-2x+4-12=0`
`⇔x^3+x^2-2x=0`
`⇔x(x^2+x-2)=0`
`⇔x(x^2+2x-x-2)=0`
`⇔x[x(x+2)-(x+2)]=0`
`⇔x(x-1)(x+2)=0`
⇔\(\left[ \begin{array}{l}x=0(N)\\x-1=0\\x+2=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0(N)\\x=1(N)\\x=-2(L)\end{array} \right.\)
Vậy `S={0;1}`
