Question

VALENDO 15 PONTOS, quem souber resolver a A, B e C

Answer 1

a) chamar (2) x = y

[ ( 2)x ] 2 - 5• ( 2 ) x + 4 = 0
y2 - 5y + 4 = 0
∆ = 25 - 16
∆ = 9
Y1 = ( 5 + 3 ) / 2
Y1 = 8 / 2
Y1 = 4
Y2 = ( 5 - 3 ) / 2
Y2 = 2 / 2
Y2 = 1

( 2 ) x = 4
(2 ) x = ( 2) 2
x = 2

(2 ) x = 1
(2) x = (2) 0
x = 0

b) ( 3 ) x • 1 / 3 + ( 3 ) x • 3 = 90
(3 ) x ( 1 / 3 + 3 ) = 90
(3 ) x • 10 / 3 = 90
(3 ) x = 90 ÷ 10 / 3
(3 ) x = 90• 3 / 10
(3 ) x = 270 / 10
( 3 ) x = 27
( 3 ) x = ( 3 ) 3
x = 3

c) chamar de (4) x = y

[ ( 4) x ] 2 + 64 = 5 • ( 4) x • 4
y2 + 64 = 5y• 4
y2 + 64 = 20y
y2 - 20y + 64 = 0
∆ = 400 - 256
∆ = 144
Y1 = ( 20 + 12 ) / 2
Y1 = 32 / 2
Y1 = 16
Y2 = ( 20 - 12 ) / 2
Y2 = 8 / 2
Y2 = 4

(4 ) x = y
(4 ) x = 1 6
( 4 ) x = ( 4 ) 2
x = 2
(4 ) x = 4
(4 ) x = ( 4 ) 1
x = 1