1.
6+6²+6³+.....+6⁹⁹
avem 99 termeni ii putem grupa cate 3
=(6+6²+6³)+.....(6⁹⁷+6⁹⁸+6⁹⁹ )=
=6(6⁰+6¹+6²)+.....+6⁹⁷(6⁰+6¹+6²)=
=6×(1+6+36)+...+6⁹⁷(1+6+36)=
=6×43+...+6⁹⁷×43=
=43(6+...+6⁹⁷) deci divizibl cu 43
2b)
2-3(x-2)-7(x+1)=3-x
2-3x+6-7x-7=3-x
1-10x=3-x
9x=-2
x=-2/9
3.
b)EC=8/2=4
DG=8+8=16
Atriunghi DEG=16×4/2=32 cm²
c)
AE=√(8²+4²)=√(64+16)=√80
EG=√(8²+4²)=√(64+16)=√80
AE+EG=√80+√80=2√80
AG=√(16²+8²)=√(256+64)=√320=2√80
rezulta ca E apartine lui AG deci A,E,G colineare
