Mangyaring lutasin ang q4 at 5?

Sagot:

# n = 0 #

Paliwanag:

Tanong 4:

Ibinigay:

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

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

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

Squaringand pagdagdag

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

# 12 = 2 (p + q) #

# p + q = 12/2 = 6 #

# p + q = 6 #

Squaring and subtracting

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

# 2sqrt11 = 4sqrt (pq) #

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

Squaring

# pq = 11/4 = 2.75 #

# x ^ 2-Sumx + Product = 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 #

Ang isa sa mga ugat ay maaaring maging p, ang iba pang magiging q.

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

Sinusunod iyan

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

Ngayon, #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#

Kaya,

# n = 0 #