Question

Calculați,folosind factorul comun:

Answer 1

a).2la a3(2la a1+2la a2+1)= 8(2+4+1)= 8×7= 56
b).3la a4(3la a2+3-11)= 81(9+3-11)= 81×1=81
c).5la a3(5la a2+5-30)= 125(25+5-30)= 125×0=0
d).2la n(2+4+8+1)= 2la n(6+8+1)= 2la n×15
e).2la n(2la a5-2la a4-2la a3)= 2la n(32-16-8)= 2la n(16-8)= 2la n×8=2la n×2la a3=2la n+3
f).3la n(5la n+2 + 3la a2×5 la n + 3×5la n+1)
g).3 la n(1+4+4la n+1 -4la n)
h).4 la n(3la n+3 + 3la n+1 + 16-9)