$1.53 \frac{\sqrt{7+4 \sqrt{3}} \cdot \sqrt{19-8 \sqrt{3}}}{4-\sqrt{3}}-\sqrt{3}=2$.
1.53 Consider the equality

$$
\frac{\sqrt{7+4 \sqrt{3}} \cdot \sqrt{19-8 \sqrt{3}}}{4-\sqrt{3}}=2+\sqrt{3}
$$

Obviously, if this equality is true, then the given equality is also true. Let

$$
a=\frac{\sqrt{7+4 \sqrt{3}} \cdot \sqrt{19-8 \sqrt{3}}}{4-\sqrt{3}}, b=2+\sqrt{3}
$$

It is easy to establish that $a>0$ and $b>0$. If in this case the equality $a^{2}=b^{2}$ holds, then $a=b$. We find

$$
\begin{aligned}
& a^{2}=\frac{(7+4 \sqrt{3})(19-8 \sqrt{3})}{(4-\sqrt{3})^{2}}= \\
& =\frac{(7+4 \sqrt{3})(19-8 \sqrt{3})}{19-8 \sqrt{3}}=7+4 \sqrt{3} \\
& b^{2}=(2+\sqrt{3})^{2}=7+4 \sqrt{3}
\end{aligned}
$$

Since $a^{2}=b^{2}$, then $a=b$, i.e., the given equality is valid.

This example can be solved faster if one guesses that both expressions under the square roots in the condition are squares of positive numbers, specifically:

$7+4 \sqrt{3}=(2+\sqrt{3})^{2}$ and $19-8 \sqrt{3}=(4-\sqrt{3})^{2}$.

Then the left side of the given equality is

$$
\frac{(2+\sqrt{3})(4-\sqrt{3})}{(4-\sqrt{3})}-\sqrt{3}=2+\sqrt{3}-\sqrt{3}=2 \text { and } 2=2
$$