Докажите, что:
1) $\sqrt{11 + 4\sqrt{7}} = \sqrt{7} + 2$;
2) $\sqrt{14 + 8\sqrt{3}} = \sqrt{8} + \sqrt{6}$.
$\sqrt{11 + 4\sqrt{7}} = \sqrt{7 + 4 + 4\sqrt{7}} = \sqrt{(\sqrt{7})^2 + 2 * 2\sqrt{7} + 2^2} = \sqrt{(\sqrt{7} + 2)^2} = |\sqrt{7} + 2| = \sqrt{7} + 2$
$\sqrt{14 + 8\sqrt{3}} = \sqrt{8 + 6 + 2 * 4\sqrt{3}} = \sqrt{(\sqrt{8})^2 + 2\sqrt{16 * 3} + (\sqrt{6})^2} = \sqrt{(\sqrt{8})^2 + 2\sqrt{48} + (\sqrt{6})^2} = \sqrt{(\sqrt{8})^2 + 2\sqrt{8 * 6} + (\sqrt{6})^2} = \sqrt{(\sqrt{8} + \sqrt{6})^2} = |\sqrt{8} + \sqrt{6}| = \sqrt{8} + \sqrt{6}$
Пожауйста, оцените решение
