Question

Efectuați,utilizând factorul comun:
a) 2 la puterea 76+2 la puterea 78+ 2 la puterea 80;
b) 3*5 la puterea 47+2*5 la puterea 48+6*5 la puterea 49;
c) 9 la puterea 1001+9 la puterea 1000+7*9 la puterea 999;
d) 7*4 la puterea 135+5*46 la puterea 136-46 la puterea 134.
Vă rog mult! E pentru mâine.

Answer 1

a) 2^76+2^78+ 2^80 =
=2^76 × (2^0+2^2+2^4)=
=2^76 × (1+4+16)=
=2^76 ×21=
=2^76 × 3 ×7


b) 3×5^47+2×5^48+6×5^49=
=5^47 × (3×5^0+2×5^1+6×5^2)=
=5^47 × (3+10+150)=
=5^47 × 163


 c) 9^1001+9^1000+7*9^999=
=9^999×(9^2+9^1+7*9^0)=
=9^999×(81+9+7)=
=9^999×97

d) 7×4^135+5×46^136-46^134 =
=7×2^270+46^134×(5×46^2-46^0)=
=7×2^270+46^134×(10580-1)=
=7×2^270+46^134×10579=
=7×2^270+2^146×23^134×71×149
=2^146(7×2^124+23^134×71×149)