Question

Pentru orice numar real x se noteaza G(x) =6x^2-x^4 si se considerea numerele a=1+√2 , b=∜2 . Aratati ca :
a) G(a) ∈ N ;
b) G(b) >6

Answer 1

$G(a)=6(1+ \sqrt{2})^2-( 1+ \sqrt{2})^4=6(1+2+2 \sqrt{2})-((1+ \sqrt{2})^2)^2 = \\ =6(3+2 \sqrt{2})-(3+2 \sqrt{2})^2=18+12 \sqrt{2}-9-12 \sqrt{2}-8= \\ =1$
1∈N

$G(b)=6 \sqrt{2}-2 \\ 6 \sqrt{2}-2\ \textgreater \ 6 \\ 6 \sqrt{2}\ \textgreater \ 2 \\ 72\ \textgreater \ 4$
deci G(b)>6

Answer 2

G(a)=G(1+√2)=6(1+√2)^2-(1+√2)^4=6(1+2√2+√2 la put a doua )-17-12√2=18+12√2-17-12√2=1
(1+√2)^2=1+2√2+(√2)^2=1+2√2+2=3+2√2
1+√2)^4={1+√2)^2}×{(1+√2^2)}=(3+2√2)(3+2√2)=9+6√2+6√2+(2√2)^2=9+12√2+8=17+12√2