1. Given positive real numbers $a, b, c$ satisfy
$$
x=\frac{a}{2 b+3 c}, y=\frac{2 b}{3 c+a}, z=\frac{3 c}{a+2 b} \text {. }
$$

Then the value of $\frac{x}{1+x}+\frac{y}{1+y}+\frac{z}{1+z}$ is ( ).
(A) 1
(B) $\frac{3}{2}$
(C) 2
(D) 3
- 1. A.

Notice that, $\frac{x}{1+x}=\frac{a}{a+2 b+3 c}$,
$$
\begin{array}{l}
\frac{y}{1+y}=\frac{2 b}{a+2 b+3 c}, \\
\frac{z}{1+z}=\frac{3 c}{a+2 b+3 c} .
\end{array}
$$

Therefore, $\frac{x}{1+x}+\frac{y}{1+y}+\frac{z}{1+z}=1$.