Prosím, vyriešte q4 a 5?

odpoveď:

# N = 0 #

vysvetlenie:

Otázka č. 4:

Vzhľadom na to:

# N = sqrt (6 + sqrt11) + sqrt (6-sqrt11) -sqrt22 #

nech

#sqrt (6 + sqrt11) = SQRTPI + sqrtq #

potom

#sqrt (6-sqrt11) = SQRTPI-sqrtq #

Squaring a pridanie

# (6 + sqrt11) + (6-sqrt11) = p + q + 2sqrt (PQ) + p + q-2sqrt (PQ) #

# 12 = 2 (p + q) #

# P + q = 12/2 = 6 #

# P + q = 6 #

Odčítavanie a odčítavanie

# (6 + sqrt11) - (6-sqrt11) = (p + q + 2sqrt (PQ)) - (p + q-2sqrt (PQ)) #=

# 2sqrt11 = 4sqrt (pq) #

#sqrt (pq) = (2sqrt11) / 4 = sqrt (11) / 2 #

kvadratúra

# PQ = 11/4 = 2,75 #

# X ^ 2-SUMX + Produkt = 0 #

# X ^ 2-6x + 2,75 = 0 #

# X ^ 2-5.5x-0,5x + 2,75 = 0 #

#X (x 5,5) -0,5 (x 5,5) = 0 #

# (X-5,5) (x 0,5) = 0 #

# X-5,5 = 0tox = 5,5 #

# X-0,5 = 0tox = 0,5 #

Jeden z koreňov môže byť p, iný bude q.

To znamená, #sqrt (6 + sqrt11) = sqrt5.5 + sqrt0.5 #

Z toho vyplýva, že

#sqrt (6-sqrt11) = sqrt5.5-sqrt0.5 #

teraz, #sqrt (6 + sqrt11) + sqrt (6-sqrt11) -sqrt22 = sqrt5.5 + sqrt0.5 + sqrt5.5-sqrt0.5-sqrt22 #

# = 2sqrt5.5-sqrt22 #

# = Qrt4sqrt5.5 = sqrt22 #

# = Sqrt (4xx5.5) -sqrt22 #

# = Sqrt22-sqrt22 #

#=0#

To znamená,

# N = 0 #