4.
AM=12√3/2=6√3
AO (inaltimea e si mediana) =(2/3) *AM=(2/3) *6√3=4√3
OM=AM-AO=6√3-4√3=2√3
as simple as that (cam pt "intelegere", nota 5-6)
5.a)
(√3-2/√3)²-(1-√3)(1+√3)= 3-2*√3*(2/√3)+4/3-(1-√3) (1+√3)=
3-4+4/3-(1-3)=3-4+4/3-1+3=
=-1+4/3-1+3=4/3+3-2=
4/3+1=4/3+3/3=7/3
medie catre GREA
b)
x+26=3x
26=2x
x=26:2
x=13
FFFFF usoara
c)
4-2x-2≤x+11
2-2x≤x+11
2-11≤x+2x
-9≤3x |:3
-3≤x
x≥-3
x∈[-3;∞)∩Z={-3;-2;-1'0;1;2.....}
x∈N∪{-3;-2;-1}
ar trebui ca pt x=-3 sa avem egalitate
verificare
4-2*(-2)≤-3+11
4+4≤8 adevarat
si ptde exemplu 0, sa se verifice ca inegalitate
4-2≤11
adevarat, bine rezolvat
nu atat grea , cat URATA!!!
