Matematyka
danior70
2017-06-25 15:30:29
Jak to rozwiązać pomozcie
Odpowiedź
invariable
2017-06-25 20:17:43

zad 6.31 a) (7 + 2√6)² = 49 + 28√6 + 24 = 73 + 28√6 (1 + √6)² = 1 + 2√6 + 6 = 7 + 2√6 b) (4 - 2√3)² = 16 - 16√3 + 12 = 28 + 16√3 = 4(7 + 4√3) (√3 - 1)² = 3 - 2√3 + 1 = 4 - 2√3 = 2(2 - √3) c) (9 - 4√5)² = 81 - 72√5+ 80 = 161 - 72√5 (√5 - 2)² = 5 - 4√5 + 4 = 9 - 4√5 d) (5 + 2√6) = 25 + 20√6 + 24 = 49 + 20√6 (√2 + √3)² = 2 + 2√6 + 3 = 5 + 2√6 zad 6.32 a) √(9 + 4√5) = √5 + 2 [√(9 + 4√5)]² = (√5 + 2)² 9 + 4√5 = 5 + 4√5 + 4 9 + 4√54 = 9 + 4√5 L = P b) √(8 + 2√7) = √7 +1 [√(8 + 2√7)]² = (√7 + 1)² 8 + 2√7 = 7 + 2√7 + 1 8 + 2√7 = 8 + 2√7 L = P zad 6.33 a) a = √6 + 2 b = 2 - √3 a > b b) a =1/(√3 - 1) = (√3 + 1)/[(√3 - 1)(√3 + 1)] = (√3 + 1)/(3 - 1) = (√3 + 1)/2 b = 1/(√5 - 1) = (√5 + 1)/[(√5 - 1)(√5 + 1)] = (√5 +1)/(5 - 1) = (√5 + 1)/4 a ≈ 1,4 b ≈ 0,8 a > b c) a = 5√2 + 2 b = 5/(√2 - 1) = 5(√2 + 1)/[(√2 - 1)(√2 + 1)] = 5(√2 + 1)/(2 - 1) = 5(√2 + 1) =  = 5√2 + 5 b > a d) a = 3√2/(√5 - √2) = 3√2(√5 + √2)/[(√5 - √2)(√5 + √2)] = 3√2(√5 + √2)/(5 - 2) =  = 3√2(√5 + √2)/3 = √2(√5 + √2) = √10 + 2 b = (√2 - 7)/√2 = √2(√2 - 7)/(√2 * √2) = (2 - 7√2)/2 = 2(1 - 3,5√2) = 1 - 3,5√2 a > b

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