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Let $a$, $b$, and $c$ be the speeds at which $A$, $B$ and $C$ work, respectively. Also, let the piece of work be worth one unit of work. Then, using the information from the problem along with basic rate formulas, we obtain the following equations: $$\begin{array}{rl}\frac{1}{a}& =m\ast \frac{1}{b+c}\\ \frac{1}{b}& =n\ast \frac{1}{a+c}\\ \frac{1}{c}& =x\ast \frac{1}{a+b}\end{array}$$ These equations can be rearranged into the following: $$\begin{array}{rl}\text{(i) }ma& =b+c\\ \text{(ii) }nb& =a+c\\ \text{(iii) }xc& =a+b\end{array}$$ Solving for $a$ in equation (i) gives us $a=\frac{b+c}{m}$. Substituting this expression for $a$ into equation (ii) yields: $$\begin{array}{rl}nb& =\frac{b+c}{m}+c\\ mnb& =b+c+mc\\ (mn-1)b& =(m+1)c\\ b& =\frac{(m+1)c}{mn-1}\end{array}$$ Finally, substituting our expressions for $a$ and $b$ into equation (iii) yields our final answer: $$\begin{array}{rl}xc& =\frac{b+c}{m}+\frac{(m+1)c}{mn-1}\\ & =\frac{\frac{(m+1)c}{mn-1}+c}{m}+\frac{(m+1)c}{mn-1}\\ & =\frac{c}{m}(\frac{m+1+mn-1}{mn-1}+\frac{m^2+m}{mn-1})\\ & =\frac{c}{m}(\frac{m^2+mn+2m}{mn-1})\\ & =c(\frac{m+n+2}{mn-1})\end{array}$$ Thus, $x=\boxed{\frac{m+n+2}{mn-1}}$.