Question

3+3x[3+3x(3+3x3)]=(3xa+3)x3+3  aflati numerul necunoscut

Answer 1

3+3x[3+3x(3+9)]=(3a+3)x3+3
3+3x(3+3x12)=9a+9+3
3+3x(3+36)=9a+12
3+3x39=9a+12
3+117=9a+12
120-12=9a
108=9a
a=108/9=12

Answer 2

3 + 3·[ 3 + 3·( 3 + 9)] = ( 3a + 3)×3 + 3
3 +3·( 3 + 3·12) = 3a·3 + 3·3 + 3
3 + 3·( 3 + 36) = 9a + 9 + 3
3 + 3 · 39 = 9a + 12
3 + 117 = 9a + 12
9a + 12 = 3 + 117
9a + 12 = 120
9a = 120 - 12
9a = 108
a = 108 : 9
a = 12
Sper ca ti-am fost de ajutor!