We are given the system of equations:

$\frac{xy}{x + y} = 1, \quad \frac{xz}{x + z} = 1, \quad \frac{yz}{y + z} = 2$

Step 1: Simplify each equation

Let’s start by simplifying the equations one by one.

From the first equation: $\frac{xy}{x+y}=1$
This implies:
$xy=x+y$
Rearranging:
$xy-x-y=0$
Factorizing:
$(x-1)(y-1)=1$

From the second equation:
$\frac{xz}{x+z}=1$
This implies:
$xz=x+z$
Rearranging:
$xz-x-z=0$
Factorizing:
$(x-1)(z-1)=1$

From the third equation:
$\frac{yz}{y+z}=2$
This implies:
$yz=2(y+z)$
Expanding:
$yz=2y+2z$
Rearranging:
$yz-2y-2z=0$
Factorizing:
$(y-2)(z-2)=4$

Step 2: Analyze the system

We now have the following system of equations:

$(x-1)(y-1)=1\\ (x-1)(z-1)=1\\ (y-2)(z-2)=4$

Step 3: Solve for the variables

From equations 1 and 2, we notice that:
$(x-1)(y-1)=(x-1)(z-1)=1$
Since both are equal to 1, we can conclude that:
$y-1=z-1$
This gives us:
$y=z$

Substituting y = z into the third equation:
$(z-2)(z-2)=4$
Expanding:
$(z-2)^2=4$
Taking the square root of both sides:
$z-2=\pm2$
Therefore,
$z=4$ or $z=0$

Step 4: Check each solution

Case 1: z = 4

If $z=4$, then $y=4$. Substituting $y=4$ into equation 1:
$(x-1)(4-1)=1\\ (x-1)(3)=1\\ x-1=\frac{1}{3}\\ x=\frac{4}{3}$

 

So, $x=\frac{4}{3}$, $y=z$, and $z=4$ is a solution.

Case 2: z = 0

 

If $z=0$, then $y=0$. Substituting $y=0$ into equation 1:
$(x-1)(0-1)=1\\ (x-1)(-1)=1\\ x-1=-1\\ x=0$

 

So, $x=0$, $y=0$, and $z=0$ is a possible solution. However, substituting into the second equation:
$\frac{0 \times 0}{0+0}\nexists$, which is not valid. So, $z=0$ is not a valid solution.

Step 5: Conclusion

The only valid solution is $x=\frac{4}{3}$, $y=4$, and $z=4$. Therefore, the value of z is: $z=4$.