1. Given $y=a x^{5}+b x^{3}+c x-5$, when $x=$ -3, $y=7$. Then, when $x=3$, the value of $y$ is ( ).
(A) -3
(B) -7
(C) -17
(D) 7
$$
-、 1 .(\mathrm{C})
$$

When $x=-3$, we get
$$
7=a(-3)^{5}+b(-3)^{3}+c(-3)-5 \text {; }
$$

When $x=3$, we get
$$
y=a(3)^{5}+b(3)^{3}+c \times 3-5 \text {; }
$$
(1) + (2), we get $7+y=-10 . \therefore y=-17$.