已知各项均为正数的等比数列 $\left\{a_{n}\right\}$, 满足 $a_{1} \cdot a_{7}=\frac{3}{4}$, 则 $a_{4}=$
$\because a_{1} \cdot a_{7}=\frac{3}{4}, \mathrm{a}_{4}{ }^{2}=\mathrm{a}_{1} \cdot \mathrm{a}_{7}, \therefore \mathrm{a}_{4}{ }^{2}=\frac{3}{4}$, 则 $\mathrm{a}_{4}= \pm \frac{\sqrt{3}}{2}$, 又 $:$ 等比数列 $\left\{\mathrm{a}_{n}\right\}$ 各项均为正数, $\therefore \mathrm{a}_{4}>0, \quad \therefore \mathrm{a}_{4}=\frac{\sqrt{3}}{2}$, 故答案是: $\frac{\sqrt{3}}{2}$.