Ha x = sqrt3 / 2, akkor {sqrt (1 + x) + sqrt (1-x)} / {sqrt (1 + x) - sqrt (1-x)}?

Ha x = sqrt3 / 2, akkor {sqrt (1 + x) + sqrt (1-x)} / {sqrt (1 + x) - sqrt (1-x)}?
Anonim

Elkezdheti a racionalizálást:

# (Sqrt (1 + x) + sqrt (1-x)) / (sqrt (1 + x) -sqrt (1-x)) × (sqrt (1 + x) + sqrt (1-x)) / (sqrt (1 + x) + sqrt (1-x)) = #

# = (Sqrt (1 + x) + sqrt (1-x)) ^ 2 / (2x) = #

# = ((1 + x) + 2sqrt (1 + x) sqrt (1-x) + (1-x)) / (2x) = #

# = (2 + 2sqrt (1-x ^ 2)) / (2x) = #

# = (1 + sqrt (1-x ^ 2)) / (x) = #

Behelyettesítve: # X = sqrt (3) / 2 # kapsz:

# = (1 + sqrt (1- 3/4)) / (sqrt (3) / 2) = (1 + 1/2) * (2 / sqrt (3)) = #

# = 3/2 * 2 / sqrt (3) = 3 / sqrt (3) = 3 / sqrt (3) * gyök (3) / sqrt (3) = sqrt (3) #

Remélem ez az, amire szüksége van!:-)