<s>[INST] Simplifying $\frac{\;\frac{x}{1-x}+\frac{1+x}{x}\;}{\frac{1-x}{x}+\frac{x}{1+x}}$ 
$$\frac{\;\;\dfrac{x}{1-x}+\dfrac{1+x}{x}\;\;}{\dfrac{1-x}{x}+\dfrac{x}{1+x}}$$

I can simplify this using the slow way ---adding the numerator and denominator, and then dividing--- but my teacher told me there is another way. Any help?
 [/INST] Multiply both the numerator and the denominator by $x(1-x)(1+x)$ and then use the fact that $(1-x)(1+x)=1-x^2$ to see that
\begin{align}
\frac{\;\;\dfrac{x}{1-x}+\dfrac{1+x}{x}\;\;}{\dfrac{1-x}{x}+\dfrac{x}{1+x}}&=\left(\frac{\;\;\dfrac{x}{1-x}+\dfrac{1+x}{x}\;\;}{\dfrac{1-x}{x}+\dfrac{x}{1+x}}\right)\frac{x(1-x)(1+x)}{x(1-x)(1+x)}
\\&=\left(\frac{x^2+(1+x)(1-x)}{(1-x)(1+x)+x^2}\right)\frac{1+x}{1-x}
\\&=\left(\frac{1}{1}\right)\frac{1+x}{1-x}
\\&=\frac{1+x}{1-x}.
\end{align}
</s>