417. Выполните действия:

Решу задания из номера 417 и 418. ### №417 а) $(2\sqrt{5} + 1)(2\sqrt{5} - 1) = (2\sqrt{5})^2 - 1^2 = 4 \cdot 5 - 1 = 20 - 1 = 19$ б) $(5\sqrt{7} - \sqrt{13})(\sqrt{13} + 5\sqrt{7}) = (5\sqrt{7})^2 - (\sqrt{13})^2 = 25 \cdot 7 - 13 = 175 - 13 = 162$ в) $(3\sqrt{2} - 2\sqrt{3})(2\sqrt{3} + 3\sqrt{2}) = (3\sqrt{2})^2 - (2\sqrt{3})^2 = 9 \cdot 2 - 4 \cdot 3 = 18 - 12 = 6$ г) $(1 + 3\sqrt{5})^2 = 1^2 + 2 \cdot 1 \cdot 3\sqrt{5} + (3\sqrt{5})^2 = 1 + 6\sqrt{5} + 9 \cdot 5 = 1 + 6\sqrt{5} + 45 = 46 + 6\sqrt{5}$ д) $(2\sqrt{3} - 7)^2 = (2\sqrt{3})^2 - 2 \cdot 2\sqrt{3} \cdot 7 + 7^2 = 4 \cdot 3 - 28\sqrt{3} + 49 = 12 - 28\sqrt{3} + 49 = 61 - 28\sqrt{3}$ е) $(2\sqrt{10} - \sqrt{2})^2 = (2\sqrt{10})^2 - 2 \cdot 2\sqrt{10} \cdot \sqrt{2} + (\sqrt{2})^2 = 4 \cdot 10 - 4\sqrt{20} + 2 = 40 - 4 \cdot 2\sqrt{5} + 2 = 42 - 8\sqrt{5}$ ### №418 а) $(\sqrt{4+\sqrt{7}} + \sqrt{4-\sqrt{7}})^2 = (\sqrt{4+\sqrt{7}})^2 + 2\sqrt{(4+\sqrt{7})(4-\sqrt{7})} + (\sqrt{4-\sqrt{7}})^2 = 4+\sqrt{7} + 2\sqrt{16-7} + 4-\sqrt{7} = 8 + 2\sqrt{9} = 8 + 2 \cdot 3 = 8 + 6 = 14$ б) $(\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}})^2 = (\sqrt{5+2\sqrt{6}})^2 - 2\sqrt{(5+2\sqrt{6})(5-2\sqrt{6})} + (\sqrt{5-2\sqrt{6}})^2 = 5+2\sqrt{6} - 2\sqrt{25 - 4 \cdot 6} + 5-2\sqrt{6} = 10 - 2\sqrt{25-24} = 10 - 2 \cdot 1 = 8$ д) $(5\sqrt{7} - 13)(5\sqrt{7} + 13) = (5\sqrt{7})^2 - 13^2 = 25 \cdot 7 - 169 = 175 - 169 = 6$ е) $(2\sqrt{2} + 3\sqrt{3})(2\sqrt{2} - 3\sqrt{3}) = (2\sqrt{2})^2 - (3\sqrt{3})^2 = 4 \cdot 2 - 9 \cdot 3 = 8 - 27 = -19$