1. Given real numbers $x, y$ satisfy $\frac{4}{x^{4}}-\frac{2}{x^{2}}=3, y^{4}+y^{2}=3$. Then the value of $\frac{4}{x^{4}}+y^{4}$ is ( ).
(A) 7
(B) 5
(C) $\frac{7+\sqrt{13}}{2}$
(D) $\frac{1+\sqrt{13}}{2}$
- 1.A.

Since $x^{2}>0, y^{2} \geqslant 0$, from the given conditions we have
$$
\begin{array}{l}
\frac{1}{x^{2}}=\frac{2+\sqrt{4+4 \times 4 \times 3}}{8}=\frac{1+\sqrt{13}}{4}, \\
y^{2}=\frac{-1+\sqrt{1+4 \times 3}}{2}=\frac{-1+\sqrt{13}}{2} . \\
\text { Therefore } \frac{4}{x^{4}}+y^{4}=\frac{2}{x^{2}}+3+3-y^{2} \\
=\frac{2}{x^{2}}-y^{2}+6=7 .
\end{array}
$$