1. If real numbers $x, y$ satisfy $4 x^{2}+y^{2}=1$, then the minimum value of $\frac{4 x y}{2 x+y-1}$ is . $\qquad$
$=.1 .1-\sqrt{2}$.  
Notice  
$$
\begin{array}{l}
\frac{4 x y}{2 x+y-1}=\frac{(2 x+y)^{2}-\left(4 x^{2}+y^{2}\right)}{2 x+y-1} \\
=\frac{(2 x+y)^{2}-1}{2 x+y-1}=2 x+y+1 .
\end{array}
$$

Let $2 x+y+1=k$. Then $y=k-2 x-1$.
Substitute into $4 x^{2}+y^{2}=1$ and rearrange to get
$$
8 x^{2}+4(1-k) x+\left(k^{2}-2 k\right)=0 \text {. }
$$

Since $x$ is a real number, we have
$$
\Delta=16(1-k)^{2}-32\left(k^{2}-2 k\right) \geqslant 0 \text {. }
$$

Solving this, we get $1-\sqrt{2} \leqslant k \leqslant 1+\sqrt{2}$.
When $x=-\frac{\sqrt{2}}{4}, y=-\frac{\sqrt{2}}{2}$, $\frac{4 x y}{2 x+y-1}$ achieves its minimum value of $1-\sqrt{2}$.