[Solution] From the given, we have $\frac{x+y}{x y}=\frac{1}{a}, \frac{x+z}{x z}=\frac{1}{b}, \frac{y+z}{y z}=\frac{1}{c}$. That is, $\frac{1}{x}+\frac{1}{y}=\frac{1}{a}, \frac{1}{x}+\frac{1}{z}=\frac{1}{b}, \frac{1}{y}+\frac{1}{z}=\frac{1}{c}$. Adding the above three equations, we get $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$. Therefore, $\frac{1}{x}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)=\frac{b c+c a-a b}{2 a b c}$. Hence, the answer is $(E)$.
1.26 If $\frac{x y}{x+y}=a, \frac{x z}{x+z}=b$ and $\frac{y z}{y+z}=c$, where $a, b$ and $c$ are all non-zero, then $x$ equals
(A) $\frac{a b c}{a b+a c+b c}$.
(B) $\frac{2 a b c}{a b+b c+a c}$.
(C) $\frac{2 a b c}{a b+a c-b c}$.
(D) $\frac{2 a b c}{a b+b c-a c}$.
(E) $\frac{2 a b c}{a c+b c-a b}$.
(2nd American High School Mathematics Examination, 1951)