Question:
If $f(x)=\frac{x-1}{x+1}$, then $f\left(\frac{1}{x}\right)+f(x)$ is equal to ______ .
Solution:
If $f(x)=\frac{x-1}{x+1}$
$f\left(\frac{1}{x}\right)=\frac{\frac{1}{x}-1}{\frac{1}{x}+1}$
$=\frac{\frac{(1-x)}{x}}{\frac{(1+x)}{x}}$
$f\left(\frac{1}{x}\right)=\frac{1-x}{1+x}$
$\therefore f\left(\frac{1}{x}\right)+f(x)=\frac{1-x}{1+x}+\frac{x-1}{x+1}$
$=\frac{1-x+x-1}{x+1}$
$\therefore f\left(\frac{1}{x}\right)+f(x)=0$