If $\dfrac{{xy}}{{x + y}} = a$ , $\dfrac{{yz}}{{y + z}} = c$, \[\dfrac{{xy}}{{x + z}} = b\], where \[a,b\] and $c$ are other than zero, then $x$ equals:
1) $\dfrac{{abc}}{{ab + ac + bc}}$
2) $\dfrac{{2abc}}{{ab + bc + ac}}$
3) $\dfrac{{2abc}}{{ab + ac - bc}}$
4) $\dfrac{{2abc}}{{ac + bc - ab}}$

Hint: The given equation $\dfrac{{xy}}{{x + y}} = a$ can be simplified as $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{a}$ . Let us assume the values $\dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z}$ as $p,q,r$ respectively and substitute these values in the given relations to form linear equations in $p,q,r$. The linear equation can be solved to find the value of $p$ . The value of $x$ will be $\dfrac{1}{p}$ .

Complete step-by-step answer:
The given relations in the question are $\dfrac{{xy}}{{x + y}} = a$ , $\dfrac{{yz}}{{y + z}} = c$ and \[\dfrac{{xy}}{{x + z}} = b\].
The equation $\dfrac{{xy}}{{x + y}} = a$ can be simplified by dividing by $xy$ in the numerator and the denominator in the L.H.S. as
\[\dfrac{{\dfrac{{xy}}{{xy}}}}{{\dfrac{x}{{xy}} + \dfrac{y}{{xy}}}} = a\]
The above equation can be simplified as
$
  \dfrac{1}{{\dfrac{1}{x} + \dfrac{1}{y}}} = a \\
   \Rightarrow \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{a} \\
$
Similarly, the equations $\dfrac{{yz}}{{y + z}} = c$ and \[\dfrac{{xy}}{{x + z}} = b\] can be written as
  $\dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{c}$
\[\dfrac{1}{x} + \dfrac{1}{z} = \dfrac{1}{b}\]
Let us assume the values $\dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z}$ as $p,q,r$ respectively .
 Substituting the values $p,q,r$ for $\dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z}$ in the above equations, we get
$p + q = \dfrac{1}{a}$, equation (1)
$q + r = \dfrac{1}{c}$, equation (2)
$p + r = \dfrac{1}{b}$, equation (3)
Adding equations 1 and 3, we get
$p + q + p + r = \dfrac{1}{a} + \dfrac{1}{b}$
Substituting the value of $q + r$ from the equation 2 in the equation $p + q + p + r = \dfrac{1}{a} + \dfrac{1}{b}$, we get
$p + p + \dfrac{1}{c} = \dfrac{1}{a} + \dfrac{1}{b}$
The above equation can be simplified as
$
  2p = \dfrac{1}{a} + \dfrac{1}{b} - \dfrac{1}{c} \\
   \Rightarrow 2p = \dfrac{{bc + ac - ab}}{{abc}} \\
   \Rightarrow p = \dfrac{{bc + ac - ab}}{{2abc}} \\
$
The value of $x$ can be found by the relation $\dfrac{1}{p} = x$
$
  x = \dfrac{1}{{\dfrac{{bc + ac - ab}}{{2abc}}}} \\
   \Rightarrow x = \dfrac{{2abc}}{{bc + ac - ab}} \\
 $
Thus option D is the correct answer.

Note: Mistakes must be avoided while dealing with fractions. Careful analysis of the given equation is necessary to formulate the way of solving the questions. The values $x,y$ and $z$ are also non zero values as $a,b$ and $c$ are non-zero values.