數學知識交流---PolyU math 5JAN2011
回答 (3)
2011-11-07 3:13 am
✔ 最佳答案
Q:√(20+√(20+√(20+…)))A:Let x=√(20+√(20+√(20+…))) x=√(20+x) x²=20+x x²-x-20=0 (x-5)(x+4)=0 x-5=0 or x+4=0 x=5 or x=-4 [Neg.,rej.]So,√(20+√(20+√(20+…)))=52011-11-07 06:44:17 補充：
Ah! Thx!!!!
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2011-11-07 4:31 am
Jason, you should firstly define whether the limit exists or not.
2011-11-07 2:53 am
設原式 = x。
x = sqrt(20+x)
x^2 = 20 + x
x^2 - x - 20 = 0
(x-5)(x+4) = 0
x1 = 5, x2 = -4(不合)
所以原式 = 5。
x = sqrt(20+x)
x^2 = 20 + x
x^2 - x - 20 = 0
(x-5)(x+4) = 0
x1 = 5, x2 = -4(不合)
所以原式 = 5。
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原文連結 [永久失效]:
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