Menjawab:
#x = 2 #
Penjelasan:
Panggilan #sqrt 49 + 20 sqrt 6 = 5 + 2 sqrt 6 = beta # kita punya
# (5 + 2 sqrt 6) ^ 1+ (5- 2 sqrt 6) ^ 1 = 10 #
untuk
#sqrt (asqrt (asqrt (a … oo)))) = 1 # dan
# x ^ 2 + x-3 - sqrt (xsqrt (xsqrt (x … oo)))) = 1 #
dan seperti itu
# a = x ^ 2-3 #
tapi
#sqrt (asqrt (asqrt (a … oo)))) = a ^ (1/2 + 1/4 + 1/8 + cdots + 1/2 ^ k + cdots) = a ^ 1 = 1 #
lalu
# 1 = x ^ 2-3 rR x = 2 #
kemudian
# x ^ 2 + x-3 - sqrt (xsqrt (xsqrt (x … oo)))) = 1 #
atau
# 1 + 2- sqrt (2sqrt (2sqrt (2 … oo)))) = 1 #
kemudian #x = 2 #