Răspuns :

Triunghiul1id

[tex]x=33\cdot3^{30}+7\cdot3^{31}+3^{33}[/tex]

[tex]x=11\cdot3\cdot3^{30}+7\cdot3^{31}+3^{33}[/tex]

[tex]x=11\cdot3^{1+30}+7\cdot3^{31}+3^{33}[/tex]

[tex]x=11\cdot3^{31}+7\cdot3^{31}+3^{33}[/tex]

[tex]x=11\cdot3^{31}+7\cdot3^{31}+3^{31}\cdot3^2[/tex]

[tex]x=3^{31}\,(11+7+3^2)[/tex]

[tex]x=3^{31}\,(11+7+9)[/tex]

[tex]x=3^{31}\,(18+9)[/tex]

[tex]x=3^{31}\cdot27[/tex]

[tex]x=3^{31}\cdot3^3=3^{31+3}[/tex]

[tex]x=3^{34} \implies x=(3^{17})^2[/tex] [tex]-patrat[/tex] [tex]perfect[/tex]

Răspuns: Ai demonstratia mai jos

Explicație pas cu pas:

Salutare!

[tex]\bf X = 33 \cdot 3^{30}+7 \cdot 3^{31}+3^{33}[/tex]

[tex]\bf X = 3^{30} \cdot (33 \cdot 3^{30-30}+7 \cdot 3^{31-30}+3^{33-30})[/tex]

[tex]\bf X = 3^{30} \cdot (33 \cdot 3^{0}+7 \cdot 3^{1}+3^{3})[/tex]

[tex]\bf X = 3^{30} \cdot (33 \cdot 1+7 \cdot 3+27)[/tex]

[tex]\bf X = 3^{30} \cdot (33 \cdot 1+21+27)[/tex]

[tex]\bf X = 3^{30} \cdot 81[/tex]

[tex]\bf X = 3^{30} \cdot 3^{4}[/tex]

[tex]\bf X = 3^{30+4}[/tex]

[tex]\bf X = 3^{34}\implies patrat \:\: perfect[/tex]

sau

[tex]\bf X = (3^{17})^{2}\implies patrat \:\: perfect[/tex]

==pav38==