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Ответ:
В этих примерах мы будем использовать формулу разности квадратов: (a-b)(a+b) = a² - b², а также формулу квадрата суммы/разности: (a±b)² = a² ± 2ab + b².
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\[ (\sqrt{17}-3)(\sqrt{17}+3) = (\sqrt{17})^2 - 3^2 = 17 - 9 = 8 \]
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\[ (\sqrt{23}-2)(\sqrt{23}+2) = (\sqrt{23})^2 - 2^2 = 23 - 4 = 19 \]
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\[ (\sqrt{47}-5)(\sqrt{47}+5) = (\sqrt{47})^2 - 5^2 = 47 - 25 = 22 \]
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\[ (\sqrt{29}-4)(\sqrt{29}+4) = (\sqrt{29})^2 - 4^2 = 29 - 16 = 13 \]
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\[ (\sqrt{41}-3)(\sqrt{41}+3) = (\sqrt{41})^2 - 3^2 = 41 - 9 = 32 \]
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\[ (\sqrt{13}-2)(\sqrt{13}+2) = (\sqrt{13})^2 - 2^2 = 13 - 4 = 9 \]
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\[ (\sqrt{7}-\sqrt{3})(\sqrt{7}+\sqrt{3}) = (\sqrt{7})^2 - (\sqrt{3})^2 = 7 - 3 = 4 \]
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\[ (\sqrt{13}-\sqrt{2})(\sqrt{13}+\sqrt{2}) = (\sqrt{13})^2 - (\sqrt{2})^2 = 13 - 2 = 11 \]
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\[ (\sqrt{17}-\sqrt{5})(\sqrt{17}+\sqrt{5}) = (\sqrt{17})^2 - (\sqrt{5})^2 = 17 - 5 = 12 \]
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\[ (\sqrt{19}-\sqrt{2})(\sqrt{19}+\sqrt{2}) = (\sqrt{19})^2 - (\sqrt{2})^2 = 19 - 2 = 17 \]
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\[ (\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3}) = (\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2 \]
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\[ (\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5}) = (\sqrt{7})^2 - (\sqrt{5})^2 = 7 - 5 = 2 \]
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\[ (\sqrt{19}-7)^2 + 14\sqrt{19} = (19 - 14\sqrt{19} + 49) + 14\sqrt{19} = 19 + 49 = 68 \]
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\[ (\sqrt{13}-3)^2 + 6\sqrt{13} = (13 - 6\sqrt{13} + 9) + 6\sqrt{13} = 13 + 9 = 22 \]
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\[ (\sqrt{11}-7)^2 + 14\sqrt{11} = (11 - 14\sqrt{11} + 49) + 14\sqrt{11} = 11 + 49 = 60 \]
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\[ (\sqrt{5}+9)^2 - 18\sqrt{5} = (5 + 18\sqrt{5} + 81) - 18\sqrt{5} = 5 + 81 = 86 \]
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\[ (\sqrt{17}+2)^2 - 4\sqrt{17} = (17 + 4\sqrt{17} + 4) - 4\sqrt{17} = 17 + 4 = 21 \]
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\[ (\sqrt{3}+8)^2 - 16\sqrt{3} = (3 + 16\sqrt{3} + 64) - 16\sqrt{3} = 3 + 64 = 67 \]
Ответ: 1) 8; 2) 19; 3) 22; 4) 13; 5) 32; 6) 9; 7) 4; 8) 11; 9) 12; 10) 17; 11) 2; 12) 2; 13) 68; 14) 22; 15) 60; 16) 86; 17) 21; 18) 67.
