20. Let $f: \mathbb{Q} \backslash\{0,1\} \rightarrow \mathbb{Q}$ be a function such that
$$
x^{2} f(x)+f\left(\frac{x-1}{x}\right)=2 x^{2}
$$
for all rational numbers $x \neq 0,1$. Here $\mathbb{Q}$ denotes the set of rational numbers. Find the value of $f\left(\frac{1}{2}\right)$.
20. Answer: 1
Substituting $x=\frac{1}{2},-1,2$, we get
$$
\begin{aligned}
\frac{1}{4} f\left(\frac{1}{2}\right)+f(-1) & =\frac{1}{2}, \\
f\left(\frac{1}{2}\right)+f(-1)+f(2) & =2, \\
4 f(2) & =8 .
\end{aligned}
$$

Solving these equations, we get $f\left(\frac{1}{2}\right)=1$. In fact the same method can be used to determine $f$. Letting $x=z, \frac{z-1}{z}, \frac{1}{1-z}$, we get
$$
\begin{array}{l}
z^{2} f(z)+f\left(\frac{z-1}{z}\right)=2 z^{2}, \\
\left(\frac{z-1}{z}\right)^{2} f\left(\frac{z-1}{z}\right)+f\left(\frac{1}{11-z}\right)=2\left(\frac{z-1}{z}\right)^{2}, \\
f(z)+\frac{1}{(1-z)^{2}} f\left(\frac{1}{1-z}\right)=\frac{z}{2(1-z)^{2}} . \\
\end{array}
$$

Using Cramer's rule, we can solve this system of linear equations in the unknowns $f(z), f\left(\frac{z-1}{z}\right), f\left(\frac{1}{1-z}\right)$. We obtain
$$
f(z)=1+\frac{1}{(1-z)^{2}}-\frac{1}{z^{2}} .
$$

Indeed one can easily check that it satisfies the given functional equation.