Molimo riješite q4 i 5?

Odgovor:

# N = 0 #

Obrazloženje:

Pitanje 4:

S obzirom na:

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

Neka, #sqrt (6 + sqrt11) = + sqrtp sqrtq #

Zatim, #sqrt (6-sqrt11) = sqrtp-sqrtq #

Kvrženje i dodavanje

# (6 + sqrt11) + (6-sqrt11) p + q + 2sqrt (PK) + p + q-2sqrt (PK) #

# 12 = 2 (p + q) #

# P + q = 12/2 = 6 #

# P + q = 6 #

Kvadratiranje i oduzimanje

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

# 2sqrt11 = 4sqrt (PK) #

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

kvadriranje

# PQ = 11/4 = 2.75 #

# X ^ 2-Sumx + Proizvodi = 0 #

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

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

Jedan od korijena može biti p, drugi će biti q.

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

Iz toga slijedi

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

Sada, #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#

Tako,

# N = 0 #