Question

gaseste 3 numere consecutive impare a căror suma este 99

Answer 1

a+a+2+a+4=99
3a+6=99
3a=99-6
3a=93
a=31
b=33
c=35
31+33+35=99

/----/  a
/----/-/  b
/----/-/-/  c

Answer 2

METODA GRAFICĂ:
• a+a+2+a+4=99 (nr. impare)
_______________
? a,a+2,a+4
_______________
a: |------|
a+2: |------|2l } 99
a+4: |------|4l
3 p.e+2+4=145
3 p.e+6=99
3 p.e=99-6
3 p.e=93
1 p.e=93:3
1 p.e=31
31+2=33 (a+2)
31+4=35 (a+4)

METODA ALGEBRICĂ:
a+a+2+a+4=99
3a+6=99
3a=99-6
3a=93
a=93:3
a=31
31+2=33 (a+2)
31+4=35 (a+4)
Verificare:
31+33+35=64+35=99 (A)
Răspuns:
a=31
a+2=33
a+4=35
Succes !
Suzuya