Let $f\left(x\right)=\frac{x}{2x-1}$  and $g\left(x\right)=\frac{x}{x-1}$ . Then the domain of the funtion h(x) = f(g(x)) + g(f(x)) is all real numbers except

The correct option is $-1,\frac{1}{2},\mathrm{and}1$.

To find the domain of the composite function $h\left(x\right)=f\left(g\right(x\left)\right)+g\left(f\right(x\left)\right)$, we must determine all values of $x$ for which both composite terms, $f\left(g\right(x\left)\right)$ and $g\left(f\right(x\left)\right)$, are defined.

Step 1: Determine the conditions for $f\left(g\right(x\left)\right)$ to be defined

1. The inner function $g\left(x\right)=\frac{x}{x-1}$ requires that its denominator is non-zero:
$x-1\ne 0\Rightarrow x\ne 1$

2. The outer function $f\left(u\right)=\frac{u}{2u-1}$ requires that its input is not equal to $\frac{1}{2}$. Therefore, we set $g\left(x\right)\ne \frac{1}{2}$:
$\frac{x}{x-1}\ne \frac{1}{2}$
$2x\ne x-1$
$x\ne -1$

Step 2: Determine the conditions for $g\left(f\right(x\left)\right)$ to be defined

1. The inner function $f\left(x\right)=\frac{x}{2x-1}$ requires that its denominator is non-zero:
$2x-1\ne 0\Rightarrow x\ne \frac{1}{2}$

2. The outer function $g\left(v\right)=\frac{v}{v-1}$ requires that its input is not equal to $1$. Therefore, we set $f\left(x\right)\ne 1$:
$\frac{x}{2x-1}\ne 1$
$x\ne 2x-1$
$x\ne 1$

Step 3: Combine all exclusions

Combining all restricted values, $x$ cannot be equal to $-1$, $\frac{1}{2}$, or $1$.

Thus, the domain of $h\left(x\right)$ consists of all real numbers except $-1,\frac{1}{2},\mathrm{and}1$.