Question

Aratati ca numarul x=33•3³⁰+7•3³¹+3³³ este pp ​

Answer 1

$x=33\cdot3^{30}+7\cdot3^{31}+3^{33}$

$x=11\cdot3\cdot3^{30}+7\cdot3^{31}+3^{33}$

$x=11\cdot3^{1+30}+7\cdot3^{31}+3^{33}$

$x=11\cdot3^{31}+7\cdot3^{31}+3^{33}$

$x=11\cdot3^{31}+7\cdot3^{31}+3^{31}\cdot3^2$

$x=3^{31}\,(11+7+3^2)$

$x=3^{31}\,(11+7+9)$

$x=3^{31}\,(18+9)$

$x=3^{31}\cdot27$

$x=3^{31}\cdot3^3=3^{31+3}$

$x=3^{34} \implies x=(3^{17})^2$ $-patrat$ $perfect$

Answer 2

Răspuns: Ai demonstratia mai jos

Explicație pas cu pas:

Salutare!

$\bf X = 33 \cdot 3^{30}+7 \cdot 3^{31}+3^{33}$

$\bf X = 3^{30} \cdot (33 \cdot 3^{30-30}+7 \cdot 3^{31-30}+3^{33-30})$

$\bf X = 3^{30} \cdot (33 \cdot 3^{0}+7 \cdot 3^{1}+3^{3})$

$\bf X = 3^{30} \cdot (33 \cdot 1+7 \cdot 3+27)$

$\bf X = 3^{30} \cdot (33 \cdot 1+21+27)$

$\bf X = 3^{30} \cdot 81$

$\bf X = 3^{30} \cdot 3^{4}$

$\bf X = 3^{30+4}$

$\bf X = 3^{34}\implies patrat \:\: perfect$

sau

$\bf X = (3^{17})^{2}\implies patrat \:\: perfect$

==pav38==