1.214. Вычислите: a) (√2-3√7)²+6√14; б) (√6-√3)²+√72; в) (√6 + √5)² - √120 - (√11)²; г) (2√5-5)² + (10+√5)²; д) (√2+1)² (3 - 2√2); е) (2-√3)²(7+4√3).

Решение:

a) \( (\sqrt{2} - 3\sqrt{7})^2 + 6\sqrt{14} = (\sqrt{2})^2 - 2 \cdot \sqrt{2} \cdot 3\sqrt{7} + (3\sqrt{7})^2 + 6\sqrt{14} = 2 - 6\sqrt{14} + 9 \cdot 7 + 6\sqrt{14} = 2 + 63 = 65 \)

б) \( (\sqrt{6} - \sqrt{3})^2 + \sqrt{72} = (\sqrt{6})^2 - 2\sqrt{6}\sqrt{3} + (\sqrt{3})^2 + \sqrt{36 \cdot 2} = 6 - 2\sqrt{18} + 3 + 6\sqrt{2} = 9 - 2\sqrt{9 \cdot 2} + 6\sqrt{2} = 9 - 2 \cdot 3\sqrt{2} + 6\sqrt{2} = 9 - 6\sqrt{2} + 6\sqrt{2} = 9 \)

в) \( (\sqrt{6} + \sqrt{5})^2 - \sqrt{120} - (\sqrt{11})^2 = (\sqrt{6})^2 + 2\sqrt{6}\sqrt{5} + (\sqrt{5})^2 - \sqrt{4 \cdot 30} - 11 = 6 + 2\sqrt{30} + 5 - 2\sqrt{30} - 11 = 11 + 2\sqrt{30} - 2\sqrt{30} - 11 = 0 \)

г) \( (2\sqrt{5} - 5)^2 + (10 + \sqrt{5})^2 = ((2\sqrt{5})^2 - 2 \cdot 2\sqrt{5} \cdot 5 + 5^2) + (10^2 + 2 \cdot 10 \cdot \sqrt{5} + (\sqrt{5})^2) = (4 \cdot 5 - 20\sqrt{5} + 25) + (100 + 20\sqrt{5} + 5) = (20 - 20\sqrt{5} + 25) + (105 + 20\sqrt{5}) = 45 - 20\sqrt{5} + 105 + 20\sqrt{5} = 150 \)

д) \( (\sqrt{2} + 1)^2 (3 - 2\sqrt{2}) = ((\sqrt{2})^2 + 2\sqrt{2} + 1^2)(3 - 2\sqrt{2}) = (2 + 2\sqrt{2} + 1)(3 - 2\sqrt{2}) = (3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 3^2 - (2\sqrt{2})^2 = 9 - (4 \cdot 2) = 9 - 8 = 1 \)

е) \( (2 - \sqrt{3})^2 (7 + 4\sqrt{3}) = (2^2 - 2 \cdot 2 \cdot \sqrt{3} + (\sqrt{3})^2)(7 + 4\sqrt{3}) = (4 - 4\sqrt{3} + 3)(7 + 4\sqrt{3}) = (7 - 4\sqrt{3})(7 + 4\sqrt{3}) = 7^2 - (4\sqrt{3})^2 = 49 - (16 \cdot 3) = 49 - 48 = 1 \)

Ответ: а) 65; б) 9; в) 0; г) 150; д) 1; е) 1.