23.25. Найдите значение выражения:

23.25. Найдите значение выражения: a) $(1-\sqrt{6})^2 + 2\sqrt{6} = 1 - 2\sqrt{6} + 6 + 2\sqrt{6} = 7$ б) $(\sqrt{2} + \sqrt{6})^2 - \sqrt{48} = 2 + 2\sqrt{12} + 6 - \sqrt{16 \cdot 3} = 8 + 4\sqrt{3} - 4\sqrt{3} = 8$ в) $(\sqrt{5} - 2)^2 + \sqrt{80} = 5 - 4\sqrt{5} + 4 + \sqrt{16 \cdot 5} = 9 - 4\sqrt{5} + 4\sqrt{5} = 9$ г) $(\sqrt{5} + \sqrt{2})^2 - \sqrt{40} = 5 + 2\sqrt{10} + 2 - \sqrt{4 \cdot 10} = 7 + 2\sqrt{10} - 2\sqrt{10} = 7$ д) $(\sqrt{3} + \sqrt{7})^2 - \sqrt{84} - (\sqrt{21})^2 = 3 + 2\sqrt{21} + 7 - \sqrt{4 \cdot 21} - 21 = 10 + 2\sqrt{21} - 2\sqrt{21} - 21 = -11$ е) $(4\sqrt{3} + 5)^2 - (10 + 2\sqrt{3})^2 = (16 \cdot 3 + 40\sqrt{3} + 25) - (100 + 40\sqrt{3} + 4 \cdot 3) = (48 + 25 + 40\sqrt{3}) - (100 + 12 + 40\sqrt{3}) = 73 + 40\sqrt{3} - 112 - 40\sqrt{3} = -39$ ж) $(2\sqrt{2} - 1)(\sqrt{8} + 1) - 15 = (2\sqrt{2} - 1)(2\sqrt{2} + 1) - 15 = (2\sqrt{2})^2 - 1^2 - 15 = 8 - 1 - 15 = 7 - 15 = -8$ з) $(3\sqrt{2} - 2)(\sqrt{18} + 2) - 17 = (3\sqrt{2} - 2)(3\sqrt{2} + 2) - 17 = (3\sqrt{2})^2 - 2^2 - 17 = 18 - 4 - 17 = 14 - 17 = -3$ и) $(\sqrt{3} - 1)^2(4 + 2\sqrt{3}) = (3 - 2\sqrt{3} + 1)(4 + 2\sqrt{3}) = (4 - 2\sqrt{3})(4 + 2\sqrt{3}) = 4^2 - (2\sqrt{3})^2 = 16 - 12 = 4$ к) $(2 + \sqrt{5})^2(9 - 4\sqrt{5}) = (4 + 4\sqrt{5} + 5)(9 - 4\sqrt{5}) = (9 + 4\sqrt{5})(9 - 4\sqrt{5}) = 9^2 - (4\sqrt{5})^2 = 81 - 80 = 1$ л) $(1 + \sqrt{2})^2(1 - \sqrt{2})^2 - 5 = ((1 + \sqrt{2})(1 - \sqrt{2}))^2 - 5 = (1^2 - (\sqrt{2})^2)^2 - 5 = (1 - 2)^2 - 5 = (-1)^2 - 5 = 1 - 5 = -4$ м) $(\sqrt{2} + 2)^2(2 - \sqrt{2})^2 - 2 = ((2 + \sqrt{2})(2 - \sqrt{2}))^2 - 2 = (2^2 - (\sqrt{2})^2)^2 - 2 = (4 - 2)^2 - 2 = 2^2 - 2 = 4 - 2 = 2$ н) $(\sqrt{3} + \sqrt{2})^2 - (\sqrt{3} - \sqrt{2})^2 - 4\sqrt{6} = (3 + 2\sqrt{6} + 2) - (3 - 2\sqrt{6} + 2) - 4\sqrt{6} = 5 + 2\sqrt{6} - 5 + 2\sqrt{6} - 4\sqrt{6} = 4\sqrt{6} - 4\sqrt{6} = 0$ о) $(\sqrt{5} + \sqrt{3})^2 - (\sqrt{3} - \sqrt{5})^2 - 4\sqrt{15} = (5 + 2\sqrt{15} + 3) - (3 - 2\sqrt{15} + 5) - 4\sqrt{15} = 8 + 2\sqrt{15} - 8 + 2\sqrt{15} - 4\sqrt{15} = 4\sqrt{15} - 4\sqrt{15} = 0$ 23.26. Избавьтесь от иррациональности в знаменателе дроби: а) $\frac{18}{\sqrt{3}} = \frac{18\sqrt{3}}{3} = 6\sqrt{3}$ б) $\frac{7}{\sqrt{14}} = \frac{7\sqrt{14}}{14} = \frac{\sqrt{14}}{2}$ в) $-\frac{\sqrt{5}}{\sqrt{35}} = -\frac{\sqrt{5} \cdot \sqrt{35}}{35} = -\frac{\sqrt{175}}{35} = -\frac{5\sqrt{7}}{35} = -\frac{\sqrt{7}}{7}$ г) $\frac{8}{5\sqrt{2}} = \frac{8\sqrt{2}}{5 \cdot 2} = \frac{8\sqrt{2}}{10} = \frac{4\sqrt{2}}{5}$ 23.27. Упростите выражение: а) $\sqrt{3} + \frac{6}{\sqrt{3}} = \sqrt{3} + \frac{6\sqrt{3}}{3} = \sqrt{3} + 2\sqrt{3} = 3\sqrt{3}$ б) $\frac{12}{\sqrt{6}} - 7\sqrt{6} = \frac{12\sqrt{6}}{6} - 7\sqrt{6} = 2\sqrt{6} - 7\sqrt{6} = -5\sqrt{6}$ в) $(\frac{14}{\sqrt{7}} + \sqrt{7}) \cdot \sqrt{7} = \frac{14}{\sqrt{7}} \cdot \sqrt{7} + \sqrt{7} \cdot \sqrt{7} = 14 + 7 = 21$