(Sqrt {10} - sqrt {2}) ^ 2 Pinasimple sa b-c sqrt {5}?

$(Sqrt {10} - sqrt {2}) ^ 2 Pinasimple sa b-c sqrt {5}?$

Sagot:

Paliwanag:

# (sqrt10 - sqrt2) ^ 2 #

# = (sqrt10 - sqrt2) (sqrt10 - sqrt2) # FOIL

# = sqrt10 * sqrt10 + sqrt10 * (- sqrt2) + sqrt10 * (- sqrt2) + (- sqrt2) * (- sqrt2) #

# = 10 - 2sqrt10sqrt2 + 2 #

# = 12 - 2sqrt10sqrt2 #

# = 12 - 2sqrt5 * sqrt2 * sqrt2 # #lArr "dahil:" sqrt10 = sqrt2 * sqrt5 #

# = 12 - 2sqrt5 * 2 #

# = 12 - 4sqrt5 #

# = 4 (3 - sqrt5) #

Paano mo pinasimple (sqrt (2) * sqrt (2)) + (sqrt (2) * -sqrt (2)) + (0 * 0)?

= 0 (sqrt2 * sqrt2) + (sqrt2 * -sqrt2) + (0 * 0) = (sqrt4) + (- sqrt4) + (0) = (2) + (- 2) + (0) = 2 - 2 + 0 = 0

Paano mo pinasimple ang 2 sqrt 3 - 4 sqrt 2 + 6 sqrt 3 + 8 sqrt 3?

22.05595867 2sqrt3 - 4sqrt2 + 6sqrt3 + 8sqrt3 =? Una, gawin ang iyong square roots sa regular na square roots: sqrt12 - sqrt32 + sqrt108 + sqrt192 =? Ipasok ito sa order na ito sa iyong calculator. Sagot: 22.05595867 Pinasimple din bilang: 16 3 - 4 2 (kredito sa Shantelle)

Paano mo pinasimple (sqrt 3 -sqrt 6) / (sqrt 3 + sqrt6)?

= -3 + 2sqrt (2) Kapag mayroon kang isang kabuuan ng dalawang square roots, ang trick ay multiply sa pamamagitan ng katumbas na pagbabawas: (sqrt (3) -sqrt (6)) / (sqrt (3) + sqrt (6) = (sqrt (3) -sqrt (6)) / (sqrt (3) + sqrt (6)) * (sqrt (3) -sqrt (6) = ((sqrt (3)) ^ 2-2 * sqrt (3) * sqrt (6) + (sqrt (6)) ^ 2) / ((sqrt (3) 2 = (3-2sqrt (18) +6) / (3-6) = (9-2 * sqrt (9 * 2)) / - 3 = (9-2 * 3sqrt (2)) / - 3 = - 3 + 2sqrt (2)