Question

x=[2^30^2*(2^6)^100*2+(64^4^100/2^899]^2+2^3007

y=5*(3^2002-3^2001-9^1000.


Aratatica numerele,,x si y'' sunt patrate perfecte...Multumesc frumos

Answer 1

x= [2^900 * 2^600 * 2 + (2^24)100 / 2^899]^2 +2^3007
x= [2^1501 +2^2400 / 2^899]^2 + 2^3007
x = (2^1501 + 2^1501)^2 + 2^3007
x= (2^1501*2)^2 + 2^3007
x=(2^1502)^2 + 2^3007
x= 2^3004 + 2^3007
x = 2^3004(1+2^3)
x= 2^3004 * 9
x = (2^1502)^2 +3^2
x = 2^1502 * 3)^2⇒  p.p.

y = 5 * (3^2002-3^2001-(3^2)^1000]
y=5*(3^2002-3^2001-3^2000)
y=5*3^2000(3^2-3^1-1)
y=5*3^2000(9-3-1)
y=5*3^2000*5
y=5^2*3^2000
y= (5*3^1000)^2⇒ p.p.