1. primul copil are [tex] \frac{x}{2}+1= \frac{x+2}{2} [/tex]
2 al doilea copil are: [tex][\frac{1}{2}(x- \frac{x+2}{2}+1)]= \frac{x+2}{4} [/tex]
3. al treilea copil are:
[tex]\frac{1}{2}(x-( \frac{x+2}{2}+ \frac{x+2}{4})]+1= \frac{x+2}{8} \\ \\ [/tex]
[tex]\frac{x+2}{2}+ \frac{x+2}{4}+ \frac{x+2}{8} =x [/tex]
aducem la acelasi numitor comun (8)
4x+8+2x+4+x+2=8x
x=14 bomboane
primul copil (x+2):2=16:2= 8 bomboane
al doilea (x+2):4=16:4= 4 bomboane
al treilea ( x+2):8 =16:2= 2 bomboane
