Question

16. Calculaţi:



P.S. : Ajutati-ma, please (pentru 50 de puncte).

Answer 1

a) [(5²)^4]^3:(16+9)^11-(0+1+1)=5^24:5^22-2=5²-2=25-2=23
b) (123+2^9+2^40):(123+2^9+2^40)=1
c) (2^8+3^4+5^6+6^4):(0+2^8+3^4+5^6+6^4)=1
d)(2.2+2.3+2.4+..........+2.4000):(2+3+4+............+4000)+2^51(2²+2+1):2^51=
la prima paranteza se da factor 2 iar ce ramane se simplifica
2+4+2+1=9
e){637+15.[15(6+12):45-6]:21}:13=
[637+15(6-6):21]:13=
637:13=49
f)[2(3+6+9+............+300)]²:(3+6+9+................+300)=2(3+6+9+...........+300)
2(3+6+9+.............+300):(1+2+3+...+100)=2.3(1+2+3+............+100):(1+2+3+..........+100)=6

Answer 2

a) [(5²)^4]^3:(16+9)^11-(0+1+1)=5^24:5^22-2=5²-2=25-2=23
b) (123+2^9+2^40):(123+2^9+2^40)=1
c) (2^8+3^4+5^6+6^4):(0+2^8+3^4+5^6+6^4)=1
d)(2.2+2.3+2.4+....+2.4000):(2+3+4+.......+4000)+2^51(2²+2+1):2^51=
ax
2+4+2+1=9
e){637+15.[15(6+12):45-6]:21}:13=
[637+15(6-6):21]:13=
637:13=45
f)[2(3+6+9+......+300)]:(3+6+9+.......+300)=2(3+6+9+......+300)
2(3+6+9+.......+300):(1+2+3+...+100)=2.3(1+2+3+......+100):(1+2+3+.....+100)=6