6) 10√3–4√48 +√27

$$10\sqrt{3} - 4\sqrt{48} + \sqrt{27} = 10\sqrt{3} - 4\sqrt{16 \cdot 3} + \sqrt{9 \cdot 3} = 10\sqrt{3} - 4 \cdot 4\sqrt{3} + 3\sqrt{3} = 10\sqrt{3} - 16\sqrt{3} + 3\sqrt{3} = (10 - 16 + 3)\sqrt{3} = -3\sqrt{3}$$ Ответ: -3√3