odpoveď:
# 1 / p ^ 2-1 / q ^ 2 = 2sqrt2 #…..# (p <q) #.
Tip: # (X-y) ^ 2 = x ^ 2 + y ^ 2-2xy = x ^ 2 + 2.xy + y ^ 2-4x #
# => (X-y) ^ 2 = (x + y) ^ 2-4x #
prosím použite '^' namiesto ' * '. # iex ^ 2 až #x ^ 2 a nie x * 2
vysvetlenie:
Myslím, že vaša kvadratická rovnica je
# 3x ^ 2-12x + 6 = 0 #.
Porovnanie s # Ax ^ 2 + bx + c = 0 #,dostaneme
# a = 3, b = -12 a c = 6 #
Ak korene tohto equn. sú #p a q #, potom
# p + q = -b / a a pq = c / a #
# t.j. + q = - (- 12) / 3 = 4 a pq = 6/3 = 2 #
teraz, # 1 / p ^ 2-1 / q ^ 2 = (q ^ 2-p ^ 2) / (p ^ 2q ^ 2) = ((q + p) (q-P)) / (PQ) ^ 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 #
korene, #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)) #
Nájsť, # 1 / p ^ 2 - 1 / q ^ 2 #
# => (1 / p + 1 / q) (1 / p-1 / q) #
# => (1 / (2 + 2sqrt (2)) + 1 / (2-2sqrt (2))) (1 / (2 + 2sqrt (2)) - 1 / (2-2sqrt (2))) #
# => (((2-2sqrt (2)) + (2 + 2sqrt (2))) / ((2-2sqrt (2)) (2 + 2sqrt (2)))) (((2-2sqrt (2)) - (2 + 2sqrt (2))) / ((2-2sqrt (2)) (2 + 2sqrt (2)))) #
# => (((2 + 2)) / ((2-2sqrt (2)) (2 + 2sqrt (2)))) (((- 2sqrt (2) -2sqrt (2))) / ((2 -2sqrt (2)) (2 + 2sqrt (2)))) #
# => ((4 (-4sqrt2)) / ((4-8)) ^ 2) #
# => ((4 (-4sqrt2)) / (- 4) ^ 2) #
# => (- sqrt2) #
