Let's calculate the following sum:

$$
\sqrt{7-\sqrt{48}}+\sqrt{5-\sqrt{24}}+\sqrt{3-\sqrt{8}}
$$
Let's try to transform the differences under the square roots into the square of a difference. We think of the difference of square roots of convenient integers, so we try to bring the values under the square root signs into the form

$$
(\sqrt{a}-\sqrt{b})^{2}=a+b-2 \sqrt{a b}=a+b-\sqrt{4 a b}
$$

For the first term, we then need $a+b=7, a b=48 / 4=12$ to hold, and the pair 3, 4 satisfies this. Since we are looking for the positive value of the square root, we have

$$
\sqrt{7-\sqrt{48}}=\sqrt{(\sqrt{4}-\sqrt{3})^{2}}=2-\sqrt{3}
$$

Similarly, for the second term, the system of equations $a+b=5, a b=6$ results; its solution is 2,3, so

$$
\sqrt{5-\sqrt{24}}=\sqrt{3}-\sqrt{2}
$$

For the third term, $a+b=3, a b=2$ has the solution 1,2, so

$$
\sqrt{3-\sqrt{8}}=\sqrt{2}-1
$$

Adding these up,

$$
\sqrt{7-\sqrt{48}}+\sqrt{5-\sqrt{24}}+\sqrt{3-\sqrt{8}}=2-\sqrt{3}+\sqrt{3}-\sqrt{2}+\sqrt{2}-1=1
$$

Remark. The above procedure can generally transform an expression of the form $\sqrt{u \pm \sqrt{v}}$ (where $u, v>0, u^{2}>v$) into a difference. In this case, the previous line of thought leads to the system of equations $a+b=u, 4 a b=v$. The roots of this system satisfy the equation $x^{2}-u x+v / 4=0$, so the values are $\left(u \pm \sqrt{u^{2}-v}\right) / 2$, leading to the identity

$$
\sqrt{u \pm \sqrt{v}}=\sqrt{\frac{u+\sqrt{u^{2}-v}}{2}} \pm \sqrt{\frac{u-\sqrt{u^{2}-v}}{2}}
$$