⚡ AZUCATION SHORTCUT:
For $h(x)$ to be defined, denominator of $f(x)$, $g(x)$, and the composite functions must not be zero.
$f(x) \to x \neq \frac{1}{2}$; $g(x) \to x \neq 1$.
Check $f(g(x)): 2g(x) - 1 \neq 0 \implies \frac{2x}{x-1} \neq 1 \implies 2x \neq x-1 \implies x \neq -1$.
Excluded values: $\mathbf{1, \frac{1}{2}, -1}$.

Step 1: Check internal constraints
The function $h(x) = f(g(x)) + g(f(x))$.
- For $g(x)$ to be defined, $x - 1 \neq 0 \implies x \neq 1$.
- For $f(x)$ to be defined, $2x - 1 \neq 0 \implies x \neq 1/2$.

Step 2: Check $f(g(x))$ domain
The denominator of $f(g(x))$ is $2g(x) - 1$. This cannot be zero:
$2\left(\frac{x}{x-1}\right) - 1 = 0 \implies \frac{2x - (x-1)}{x-1} = 0$
$x + 1 = 0 \implies x = -1$.
So, $x \neq -1$.

Step 3: Check $g(f(x))$ domain
The denominator of $g(f(x))$ is $f(x) - 1$. This cannot be zero:
$\frac{x}{2x-1} - 1 = 0 \implies \frac{x - (2x-1)}{2x-1} = 0$
$-x + 1 = 0 \implies x = 1$.
We already found $x \neq 1$ in Step 1.

Final Answer:
The values to exclude are $1, 1/2,$ and $-1$.
Correct Option: A.

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