Question

cum se rezova aceste exercitii ,va rog

Answer 1

c)
[(9+18+27+...+900)-(7+14+21+...+700)]:(1+2+3+...+100)=
=[9(1+2+3+...+100)-7(1+2+3+...+100)]:(1+2+3+...+100)=
=[(1+2+3+...+100)(9-7)]:(1+2+3+...+100)=
=9-7=2

2.
2^77 + 2^76 + 2^75 + 2^74=
=2^74( 2^3 + 2^2 + 2^1 + 1)=
=2^74 (8+4+2+1)=
=2^74 ·15
=>2^77 + 2^76 + 2^75 + 2^74 divizibil cu 15

3.
a)

a=7+7-21:3
a=14-7
a=7

b=9-3·[9-3·(3-3:3)]
b=9-3·[9-3·(3-1)]
b=9-3·(9-3·2)
b=9-3·(9-6)
b=9-3·3
b=9-9
b=0

c=[(6^0 ·8^2-2^5):2^3+1]·2-9
c=[(1·64-32):8+1]·2-9
c=(32:8+1)·2-9
c=(4+1)·2-9
c=5·2-9
c=10-9
c=1


b)a=7
a nu e difizibil cu 5
Propozitia este falsa

c)
c^2013 +b^a=
=1^2013 + 0^7=
=1+0=
=1