## Task B-1.1.

If $x y z=a b c$, what is $\frac{b x}{x y+a b+b x}+\frac{c y}{y z+b c+c y}+\frac{a z}{z x+c a+a z}$?
## Solution.

First, extend the first fraction by $z$, and the second by $a$.

$$
\begin{aligned}
& \frac{b x}{x y+a b+b x} \cdot \frac{z}{z}+\frac{c y}{y z+b c+c y} \cdot \frac{a}{a}+\frac{a z}{z x+c a+a z}= \\
& \frac{b x z}{x y z+a b z+b x z}+\frac{c a y}{a y z+a b c+a c y}+\frac{a z}{z x+c a+a z}
\end{aligned}
$$

Replace the expression $xyz$ in the denominator of the first fraction with $abc$, and the expression $abc$ in the denominator of the second fraction with $xyz$.

$$
\begin{gathered}
\frac{b x z}{a b c+a b z+b x z}+\frac{a c y}{a y z+x y z+a c y}+\frac{a z}{z x+c a+a z}= \\
\frac{b x z}{b(a c+a z+x z)}+\frac{a c y}{y(a z+x z+a c)}+\frac{a z}{z x+c a+a z}=\frac{x z+a c+a z}{a c+x z+a z}=1 .
\end{gathered}
$$