Решение:
- а) \(7\sqrt{3} - \sqrt{48} + \sqrt{27} = 7\sqrt{3} - \sqrt{16 \cdot 3} + \sqrt{9 \cdot 3} = 7\sqrt{3} - 4\sqrt{3} + 3\sqrt{3} = (7-4+3)\sqrt{3} = 6\sqrt{3}\)
- б) \(\sqrt{2}(\sqrt{8} + 4\sqrt{2}) = \sqrt{2} \cdot \sqrt{8} + \sqrt{2} \cdot 4\sqrt{2} = \sqrt{16} + 4 \cdot 2 = 4 + 8 = 12\)
- в) \((\sqrt{3} + 5)^2 = (\sqrt{3})^2 + 2 \cdot \sqrt{3} \cdot 5 + 5^2 = 3 + 10\sqrt{3} + 25 = 28 + 10\sqrt{3}\)
- г) \((\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3}) = (\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2\)
Ответ: а) \(6\sqrt{3}\); б) 12; в) \(28 + 10\sqrt{3}\); г) 2.