bjr
1)
2x+3x+2x-10-5x = (on associe les termes en x)
(2x + 3x + 2x - 5x) - 10 =
x(2 + 3 + 2 - 5) - 10 =
x*(2) - 10 =
2x - 10
2)
33x-22x-2x+5-2 = (33x - 22x - 2x) + 5 - 2
= x(33 - 22 - 2) + 3
= x(33 - 24) + 3
= x*(9) + 3
= 9x + 3
3)
5x-2x+3+5-2x-5 = 5x - 2x - 2x + 3 + 5 - 5
= (5x - 2x - 2x ) + 3
= x(5 - 2 - 2) + 3
= x*(1) + 3
= x + 3
4)
2x(3x + 5x) + 2 = 2x(3x + 5x) + 2
= 2x*3x + 2x*5x + 2 on multiplie par 2x les termes
entre parenthèses
= 6x² + 10x² + 2 [2x*3x = 2*3*x*x = 6x² ]
on associe les terme en x²
= (6x² + 10x²) + 2
= x²(6 + 10) + 2
= 16x² + 2
5)
3x(2 + 3 -5) = 3x*(0) = 0
la somme entre parenthèses est nulle
