Упрости выражение: a) √75 + √48-√300

a) $\sqrt{75} + \sqrt{48} - \sqrt{300} = \sqrt{25 \cdot 3} + \sqrt{16 \cdot 3} - \sqrt{100 \cdot 3} = 5\sqrt{3} + 4\sqrt{3} - 10\sqrt{3} = (5+4-10)\sqrt{3} = -1\sqrt{3} = -\sqrt{3}$. б) $3\sqrt{8} - \sqrt{50} + 2\sqrt{18} = 3\sqrt{4 \cdot 2} - \sqrt{25 \cdot 2} + 2\sqrt{9 \cdot 2} = 3 \cdot 2\sqrt{2} - 5\sqrt{2} + 2 \cdot 3\sqrt{2} = 6\sqrt{2} - 5\sqrt{2} + 6\sqrt{2} = (6 - 5 + 6)\sqrt{2} = 7\sqrt{2}$. в) $\sqrt{242} - \sqrt{200} + \sqrt{8} = \sqrt{121 \cdot 2} - \sqrt{100 \cdot 2} + \sqrt{4 \cdot 2} = 11\sqrt{2} - 10\sqrt{2} + 2\sqrt{2} = (11 - 10 + 2)\sqrt{2} = 3\sqrt{2}$. г) $\sqrt{75} - 0{,}1\sqrt{300} - \sqrt{27} = \sqrt{25 \cdot 3} - 0{,}1\sqrt{100 \cdot 3} - \sqrt{9 \cdot 3} = 5\sqrt{3} - 0{,}1 \cdot 10\sqrt{3} - 3\sqrt{3} = 5\sqrt{3} - \sqrt{3} - 3\sqrt{3} = (5 - 1 - 3)\sqrt{3} = 1\sqrt{3} = \sqrt{3}$. д) $\sqrt{98} - \sqrt{72} + 0{,}5\sqrt{8} = \sqrt{49 \cdot 2} - \sqrt{36 \cdot 2} + 0{,}5\sqrt{4 \cdot 2} = 7\sqrt{2} - 6\sqrt{2} + 0{,}5 \cdot 2\sqrt{2} = 7\sqrt{2} - 6\sqrt{2} + \sqrt{2} = (7 - 6 + 1)\sqrt{2} = 2\sqrt{2}$. **Ответ:** a) $-\sqrt{3}$; б) $7\sqrt{2}$; в) $3\sqrt{2}$; г) $\sqrt{3}$; д) $2\sqrt{2}$.