4. Find all real numbers $x$ that satisfies
$$
\frac{x^{4}+x+1}{x^{4}+x^{2}-x-4}=\frac{x^{4}+1}{x^{4}+x^{2}-4} .
$$
Answer: $-1,0,1$
Solution: Let $f(x)=x^{4}+x+1$ and $g(x)=x^{4}+x^{2}-x-4$. The equation is then equivalent to
$$
\frac{f(x)}{g(x)}=\frac{f(x)-x}{g(x)+x} \Longleftrightarrow x(f(x)+g(x))=0 .
$$

Hence, $x=0$ or $f(x)+g(x)=2 x^{4}+x^{2}-3=\left(2 x^{2}+3\right)\left(x^{2}-1\right)=0$ while gives $-1,0,1$ as acceptable values of $x$.