$\sqrt{\frac{b + 1}{2} - \sqrt{b}} - \sqrt{\frac{b + 1}{2} + \sqrt{b}} = \sqrt{\frac{b - 2\sqrt{b} + 1}{2}} - \sqrt{\frac{b + 2\sqrt{b} + 1}{2}} = \sqrt{\frac{(\sqrt{b} - 1)^2}{2}} - \sqrt{\frac{(\sqrt{b} + 1)^2}{2}} = \frac{|\sqrt{b} - 1|}{\sqrt{2}} - \frac{|\sqrt{b} + 1|}{\sqrt{2}}$
Так как b ≥ 1, $|\sqrt{b} - 1| = \sqrt{b} - 1, |\sqrt{b} + 1| = \sqrt{b} + 1$
Получаем:
$\frac{\sqrt{b} - 1}{\sqrt{2}} - \frac{\sqrt{b} + 1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$

$\sqrt{\frac{c + 4}{4} + \sqrt{c}} - \sqrt{\frac{c + 4}{4} - \sqrt{c}} = \sqrt{\frac{c + 4\sqrt{c} + 4}{4}} - \sqrt{\frac{c - 4\sqrt{c} + 4}{4}} = \frac{\sqrt{(\sqrt{c} + 2)^2}}{2} - \frac{\sqrt{(\sqrt{c} - 2)^2}}{2} = \frac{|\sqrt{c} + 2|}{2} - \frac{|\sqrt{c} - 2|}{2}$
Так как c ≥ 4, $|\sqrt{c} - 2| = \sqrt{c} - 2, |\sqrt{c} + 2| = \sqrt{c} + 2$
Получаем:
$\frac{\sqrt{c} + 2}{2} - \frac{\sqrt{c} - 2}{2} = \frac{4}{2} = 2$