(xqqrt (x ... o) + (5 + 2sqrt6) ^ (x ^ 2 + x-3 - sqrt (xsqrt (xsqrt (x ...) )) = 10 = a = x ^ 2-3, ולאחר מכן x?

(xqqrt (x ... o) + (5 + 2sqrt6) ^ (x ^ 2 + x-3 - sqrt (xsqrt (xsqrt (x ...) )) = 10 = a = x ^ 2-3, ולאחר מכן x?
Anonim

תשובה:

#x = 2 #

הסבר:

מתקשר #sqrt 49 + 20 sqrt 6 = 5 + 2 sqrt 6 = beta # יש לנו

# 5 + 2 sqrt 6) ^ 1 + (5 - 2 sqrt 6) ^ 1 = 10 #

ל

#sqrt (asqrt (asqrt (a … oo)) = 1 # ו

# x ^ 2 + x-3 - sqrt (xsqrt (xsqrt (x … oo)) = 1 #

וכך

# a = x ^ 2-3 #

אבל

# asqrt (asqrt (a …)) a = (1/2 + 1/4 + 1/8 + cdots + 1/2 ^ k + cdots) = a = 1 = 1 #

ואז

# 1 = x ^ 2-3 rRrr x = 2 #

לאחר מכן

# x ^ 2 + x-3 - sqrt (xsqrt (xsqrt (x … oo)) = 1 #

או

# 1 + 2 sqrt (2sqrt (2sqrt (2 … oo))) = 1 #

לאחר מכן #x = 2 #