1)
1+3×{[17+5×[730-6×(x-3×32)]}+1=2003
1+3×{[17+5×[730-6×(x-96)]} = 2003-1=2002
3×{[17+5×[730-6×(x-96)]} = 2002-1 = 2001
17+5×[730-6×(x-96)] = 2001 : 3 = 667
5×[730-6×(x-96)] = 667 - 17 =650
730 - 6×(x-96) = 650:5 = 130
6×(x-96) = 730 - 130 = 600
x - 96 = 600 : 6 = 100
x = 100 + 96
x = 196
Al doilea exercitiu nu este scris corect in totalitate:
o paranteza mica lipseste si 3121:12 = da cu rest.
