5. If $\log _{4}(x+2 y)+\log _{4}(x-2 y)=1$, then the minimum value of $|x|-|y|$ is

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5. $\sqrt{3}$ Detailed Explanation: $\left\{\begin{array}{l}x+2 y>0 \\ x-2 y>0 \\ (x+2 y)(x-2 y)=4\end{array} \Rightarrow\left\{\begin{array}{l}x>2|y| \\ x^{2}-4 y^{2}=4\end{array}\right.\right.$.

By symmetry, only consider $y \geqslant 0$, since $x>0$, therefore, we only need to find the minimum value of $x-y$. Let $x-y=u$, substitute into $x^{2}-4 y^{2}=4$, we get $3 y^{2}-2 u y+4-u^{2}=0$. This quadratic equation in $y$ clearly has real roots, hence $\Delta=16\left(u^{2}-3\right) \geqslant 0$.