(sqrt (49 + 20sqrt6)) ^ (sqrt (asqrt (a ... oo) + (5-2sqrt6) ^ (x ^ 2 + x-3 - sqrt (xsqrt (xsqrt (x ... oo)) ))) = 10 wobei a = x ^ 2-3, dann ist x?

Antworten:

#x = 2 #

Berufung #sqrt 49 + 20 sqrt 6 = 5 + 2 sqrt 6 = beta # wir haben

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

zum

#sqrt (asqrt (asqrt (a … oo))) = 1 # und

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

und so weiter

# a = x ^ 2-3 #

aber

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

und dann

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

dann

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

oder

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

dann #x = 2 #