Jika g (x) = root [3] {x ^ {2} - 1} + 2 sqrt {x + 1}, apakah g (3)?

Jawapan:

#g (3) = 6 #

Penjelasan:

Hanya pengganti 3 di mana sahaja ada # x #

#g (3) = root (3) (3 ^ 2-1) + 2sqrt (3 + 1) #

#g (3) = root (3) 8 + 2sqrt4 #

#g (3) = 2 + 2sqrt4 #

#g (3) = 2 + 2xx2 #

#g (3) = 2 + 4 #

#g (3) = 6 #

Jawapan:

#g (3) = 6 #

Penjelasan:

Palam #3# dalam untuk # x #

#g (3) = root3 (3 ^ 2-1) + 2sqrt (3 + 1) #

Mudahkan

#g (3) = root3 (9-1) + 2sqrt (4) #

#g (3) = root3 (8) + 2sqrt (4) #

#g (3) = 2 + 2 (2) #

#g (3) = 2 + 4 #

#g (3) = 6 #