Kaj je (sqrt (5+) sqrt (3)) / (sqrt (3+) sqrt (3+) sqrt (5)) - (sqrt (5-) sqrt (3)) / (sqrt (3+) sqrt (3-) sqrt (5))?

Kaj je (sqrt (5+) sqrt (3)) / (sqrt (3+) sqrt (3+) sqrt (5)) - (sqrt (5-) sqrt (3)) / (sqrt (3+) sqrt (3-) sqrt (5))?
Anonim

Odgovor:

#2/7#

Pojasnilo:

Vzamemo, # A = (sqrt5 + sqrt3) / (sqrt3 + sqrt3 + sqrt5) - (sqrt5-sqrt3) / (sqrt3 + sqrt3-sqrt5) #

# = (sqrt5 + sqrt3) / (2sqrt3 + sqrt5) - (sqrt5-sqrt3) / (2sqrt3-sqrt5) #

# = (sqrt5 + sqrt3) / (2sqrt3 + sqrt5) - (sqrt5-sqrt3) / (2sqrt3-sqrt5) #

# = ((sqrt5 + sqrt3) (2sqrt3-sqrt5) - (sqrt5-sqrt3) (2sqrt3 + sqrt5)) / ((2sqrt3 + sqrt5) (2sqrt3-sqrt5) #

# = ((2sqrt15-5 + 2 * 3-sqrt15) - (2sqrt15 + 5-2 * 3-sqrt15)) / ((2sqrt3) ^ 2- (sqrt5) ^ 2) #

# = (prekliči (2sqrt15) -5 + 2 * 3zaključi (-sqrt15) - prekliči (2sqrt15) -5 + 2 * 3 + prekliči (sqrt15)) / (12-5) #

#=(-10+12)/7#

#=2/7#

Upoštevajte, da, če so v imenovalec

# (sqrt3 + sqrt (3 + sqrt5)) in (sqrt3 + sqrt (3-sqrt5)) #

potem bo odgovor spremenjen.