Упрости выражение √75 + √48 - √300

Задание 421. 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) $\sqrt{75} - 0.1\sqrt{300} - \sqrt{27} = \sqrt{25 \cdot 3} - 0.1\sqrt{100 \cdot 3} - \sqrt{9 \cdot 3} = 5\sqrt{3} - 0.1 \cdot 10 \sqrt{3} - 3\sqrt{3} = 5\sqrt{3} - 1\sqrt{3} - 3\sqrt{3} = (5 - 1 - 3)\sqrt{3} = 1\sqrt{3} = \sqrt{3}$ д) $\sqrt{98} - \sqrt{72} + 0.5\sqrt{8} = \sqrt{49 \cdot 2} - \sqrt{36 \cdot 2} + 0.5 \sqrt{4 \cdot 2} = 7\sqrt{2} - 6\sqrt{2} + 0.5 \cdot 2 \sqrt{2} = 7\sqrt{2} - 6\sqrt{2} + 1\sqrt{2} = (7 - 6 + 1)\sqrt{2} = 2\sqrt{2}$ Задание 422. Допущение: в задании 422а пропущен знак умножения: $\sqrt{8p} - \sqrt{2p} + \sqrt{18p}$ a) $\sqrt{8p} - \sqrt{2p} + \sqrt{18p} = \sqrt{4 \cdot 2p} - \sqrt{2p} + \sqrt{9 \cdot 2p} = 2\sqrt{2p} - \sqrt{2p} + 3\sqrt{2p} = (2 - 1 + 3)\sqrt{2p} = 4\sqrt{2p}$ б) $\sqrt{160c} + 2\sqrt{40c} - 3\sqrt{90c} = \sqrt{16 \cdot 10c} + 2\sqrt{4 \cdot 10c} - 3\sqrt{9 \cdot 10c} = 4\sqrt{10c} + 2 \cdot 2 \sqrt{10c} - 3 \cdot 3 \sqrt{10c} = 4\sqrt{10c} + 4\sqrt{10c} - 9\sqrt{10c} = (4 + 4 - 9)\sqrt{10c} = -1\sqrt{10c} = -\sqrt{10c}$ в) $5\sqrt{27m} - 4\sqrt{48m} - 2\sqrt{12m} = 5\sqrt{9 \cdot 3m} - 4\sqrt{16 \cdot 3m} - 2\sqrt{4 \cdot 3m} = 5 \cdot 3 \sqrt{3m} - 4 \cdot 4 \sqrt{3m} - 2 \cdot 2 \sqrt{3m} = 15\sqrt{3m} - 16\sqrt{3m} - 4\sqrt{3m} = (15 - 16 - 4)\sqrt{3m} = -5\sqrt{3m}$ г) $\sqrt{54} - \sqrt{24} + \sqrt{150} = \sqrt{9 \cdot 6} - \sqrt{4 \cdot 6} + \sqrt{25 \cdot 6} = 3\sqrt{6} - 2\sqrt{6} + 5\sqrt{6} = (3 - 2 + 5)\sqrt{6} = 6\sqrt{6}$ д) $3\sqrt{2} + \sqrt{32} - \sqrt{200} = 3\sqrt{2} + \sqrt{16 \cdot 2} - \sqrt{100 \cdot 2} = 3\sqrt{2} + 4\sqrt{2} - 10\sqrt{2} = (3 + 4 - 10)\sqrt{2} = -3\sqrt{2}$ e) $2\sqrt{72} - \sqrt{50} - 2\sqrt{8} = 2\sqrt{36 \cdot 2} - \sqrt{25 \cdot 2} - 2\sqrt{4 \cdot 2} = 2 \cdot 6 \sqrt{2} - 5\sqrt{2} - 2 \cdot 2 \sqrt{2} = 12\sqrt{2} - 5\sqrt{2} - 4\sqrt{2} = (12 - 5 - 4)\sqrt{2} = 3\sqrt{2}$