Question

Sa se afle termenul necunoscut din egalitatea:
3+3x[3+3x(3+3x3)]=(3xa+3)x3+3

Answer 1

Răspuns:

3 + 3 × (3 + 3 × 12) = 9a + 9 + 3

3 + 3 × 39 = 9a + 12

3 + 117 - 12 = 9a

108 = 9a

a = 108 : 9

a = 12

Answer 2

Răspuns: $\green{\underline{~\bf a = 12~}}$  

Explicație pas cu pas:

$\bf 3+3\cdot [3+3\cdot(3+3\cdot3)]=(3\cdot a+3)\cdot3+3$

$\bf 3+3\cdot [3+3\cdot(3+9)]=9a+9+3$

$\bf 3+3\cdot (3+3\cdot12)=9a+9+3$

$\bf 3+3\cdot (3+36)=9a+12$

$\bf 3+3\cdot 39 =9a+12$

$\bf 3+117 =9a+12$

$\bf 120 =9a+12$

$\bf 120 -12 =9a$

$\bf 108 =9a$

$\bf 9a = 108 ~~~\bigg|:9$

$\green{\underline{~\bf a = 12~}}$

$==pav38==$