Odgovor:
# 1 / p ^ 2-1 / q ^ 2-2sqrt2 #…..# (p <q) #.
Savjet: # (X-y) ^ 2-x ^ 2 + y ^ 2-2xy = x ^ 2 + y + 2xy ^ 2-4xy #
# => (X-y) ^ 2 = (x + y) ^ 2-4xy #
koristite '^' umjesto ' * '. # i.e.x ^ 2 to #x ^ 2, a ne x * 2
Obrazloženje:
Mislim da je vaša kvadratna jednadžba
# 3x ^ 2-12x + 6 = 0 #.
U usporedbi s # X ^ 2 + bx + c = 0 #, dobivamo
# a = 3, b = -12 i c = 6 #
Ako korijeni ove equn. su #p i q #, onda
# p + q = -b / a i pq = c / a #
# i.e.p + q = - (- 12) / 3 = 4 i pq = 6/3 = 2 #
Sada, # 1 / p ^ 2-1 / q ^ 2 = (q ^ 2-p ^ 2) / (p ^ 2q ^ 2) = ((q + p) (q-p)) / (PK) ^ 2 #,….# (p <q) #
# => 1 / p ^ 2-1 / q ^ 2 = ((4) sqrt ((q-p) ^ 2),) / 2 ^ 2-sqrt ((q-p) ^ 2 #
# => 1 / p ^ 2-1 / q ^ 2-sqrt ((q + p) ^ 2-4pq) = sqrt (4 ^ 2-4 (2) #
# => 1 / p ^ 2-1 / q ^ 2-sqrt (16-8) = sqrt8 = 2sqrt2 #….# (p <q) #
# 3x ^ 2-12x + 6 = 0 #
# => x ^ 2 - 4x + 2 = 0 #
Korijenje, #x = (- b + -sqrt (b ^ 2-4ac)) / (2a) #
# X = (4 + -sqrt (16-4 * 1 * 2)) / (2) #
# x = (4 + -sqrt (8)) / (2) = (4 + 2sqrt (2)) / (2) #
# X = (2 + -2sqrt (2)) *
Pronaći, # 1 / p ^ 2 - 1 / q ^ 2 #
# => (1 / p + 1 / q) (1 / p-1 / q) #
# => (1 / (2 + 2sqrt (2)) + 1 / (2-2sqrt (2)) i) (1 / (2 + 2sqrt (2)) - 1 / (2-2sqrt (2))) *
# => (((2-2sqrt (2)) + (2 + 2sqrt (2))) / ((2-2sqrt (2)), (2 + 2sqrt (2)) i)) (((2-2sqrt (2)) - (2 + 2sqrt (2))) / ((2-2sqrt (2)), (2 + 2sqrt (2)) i)) *
# => (((2 + 2)) / ((2-2sqrt (2)), (2 + 2sqrt (2)) i)) (((- 2sqrt (2) -2sqrt (2))) / ((2 -2sqrt (2)), (2 + 2sqrt (2)) i)) *
# => ((4 (-4sqrt2)) / ((4-8)) ^ 2) *
# => ((4 (-4sqrt2)) / (- 4) ^ 2) *
# => (- sqrt2) #
