$r)\\\sqrt{2}\sqrt{4+\sqrt{15}}\\=\sqrt{2(4+\sqrt{15})}\\=\sqrt{8+2\sqrt{15}}\\=\sqrt{8+2.\sqrt{5}.\sqrt{3}}\\=\sqrt{(\sqrt{5}+\sqrt{3})^2}\\=\sqrt{5}+\sqrt{3}\\t)\\\sqrt{227-30\sqrt{2}}\\=\sqrt{227-2.15.\sqrt{2}}\\=\sqrt{(15-\sqrt{2})^2}\\=|15-\sqrt{2}|\\=15-\sqrt{2}\\u)\\\sqrt{\dfrac{59}{25}+\dfrac{6}{5}\sqrt{2}}\\=\sqrt{\dfrac{59}{25}+2.\dfrac{3}{5}.\sqrt{2}}\\=\sqrt{\Big(\dfrac{3}{5}+\sqrt{2}\Big)^2}\\=\dfrac{3}{5}+\sqrt{2}\\v)\\\sqrt{2}\sqrt{8+3\sqrt{7}}\\=\sqrt{2(8+3\sqrt{7})}\\=\sqrt{16+6\sqrt{7}}\\=\sqrt{16+2.3.\sqrt{7}}\\=\sqrt{(3+\sqrt{7})^2}\\=3+\sqrt{7}\\w)\\\sqrt{123+22\sqrt{2}}\\=\sqrt{123+2.11.\sqrt{2}}\\=\sqrt{(11+\sqrt{2})^2}\\=11+\sqrt{2}\\x)\\\sqrt{10-2\sqrt{21}}\\=\sqrt{10-2.\sqrt{3}.\sqrt{7}}\\=\sqrt{(\sqrt{7}-\sqrt{3})^2}\\=|\sqrt{7}-\sqrt{3}|\\=\sqrt{7}-\sqrt{3}\\c')\\\sqrt{11+6\sqrt{2}}\\=\sqrt{11+2.3.\sqrt{2}}\\=\sqrt{(3+\sqrt{2})^2}\\=3+\sqrt{2}\\d')\\\sqrt{12-3\sqrt{7}}\\=\dfrac{2\sqrt{12-3\sqrt{7}}}{2}\\=\dfrac{\sqrt{48-12\sqrt{7}}}{2}\\=\dfrac{\sqrt{48-2.\sqrt{6}.\sqrt{6}.\sqrt{7}}}{2}\\=\dfrac{\sqrt{48-2.\sqrt{6}.\sqrt{42}}}{2}\\=\dfrac{\sqrt{(\sqrt{42}-\sqrt{6})^2}}{2}\\=\dfrac{|\sqrt{42}-\sqrt{6}|}{2}\\=\dfrac{\sqrt{42}-\sqrt{6}}{2}\\e')\\\sqrt{2-\sqrt{3}}\\=\dfrac{\sqrt{2}\sqrt{2-\sqrt{3}}}{\sqrt{2}}\\=\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\\=\dfrac{\sqrt{4-2.1.\sqrt{3}}}{\sqrt{2}}\\=\dfrac{\sqrt{(\sqrt{3}-1)^2}}{\sqrt{2}}\\=\dfrac{|\sqrt{3}-1|}{\sqrt{2}}\\=\dfrac{\sqrt{3}-1}{\sqrt{2}}\\=\dfrac{\sqrt{2}(\sqrt{3}-1)}{2}\\=\dfrac{\sqrt{6}-\sqrt{2}}{2}\\f')\\\sqrt{3-\sqrt{5}}\\=\dfrac{\sqrt{2}\sqrt{3-\sqrt{5}}}{\sqrt{2}}\\=\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\\=\dfrac{\sqrt{6-2.1.\sqrt{5}}}{\sqrt{2}}\\=\dfrac{\sqrt{(\sqrt{5}-1)^2}}{\sqrt{2}}\\=\dfrac{|\sqrt{5}-1|}{\sqrt{2}}\\=\dfrac{\sqrt{5}-1}{\sqrt{2}}\\=\dfrac{\sqrt{2}(\sqrt{5}-1)}{2}\\=\dfrac{\sqrt{10}-\sqrt{2}}{2}\\z)\\\sqrt{2-\sqrt{3}}.\,(\sqrt{6}+\sqrt{2})\\=\sqrt{2}\sqrt{2-\sqrt{3}}.\,(\sqrt{3}+1)\\=\sqrt{4-2\sqrt{3}}.\,(\sqrt{3}+1)\\=\sqrt{4-2.1.\sqrt{3}}.\,(\sqrt{3}+1)\\=\sqrt{(\sqrt{3}-1)^2}.\,(\sqrt{3}+1)\\=|\sqrt{3}-1|(\sqrt{3}+1)\\=(\sqrt{3}-1)(\sqrt{3}+1)\\=(\sqrt{3})^2-1\\=3-1\\=2$