Question

Se consideră numărul n={(2¹⁴-2¹³+2¹⁴)³:[43-(1008:18+2³)²:4⁴]+2³⁶}:4¹⁷ a) Arătați ca numarul natural n este cub perfect. b)Câte cifre are numarul natural n¹⁰
repede dau coroană​

Answer 1

Răspuns:

$n = \big\{\big(2^{14}-2^{13} + 2^{12}\big)^{3} : \big[43 - \big(1008 : 18 + 2^{3}\big)^{2} : 4^{4}\big] + 2^{36}\big\} : 4^{17}$

$n = \big\{ \big[2^{12}\cdot\big(2^{14-12}-2^{13-12} + 2^{12-12} \big)\big]^{3} : \big[43 - \big(56+ 8\big)^{2} : 4^{4}\big] + 2^{36}\big\} : \big(2^{2}\big)^{17}$

$n = \big\{ \big[2^{12}\cdot\big(2^{2}-2^{1} + 2^{0} \big)\big]^{3} : \big(43 - 64^{2} : 4^{4}\big) + 2^{36}\big\} :2^{2\cdot17}$

$n = \big\{ \big[2^{12}\cdot\big(4-2 + 1\big)\big]^{3} : \big[43 - \big(4^4 \big)^{2} : 4^{4}\big] + 2^{36}\big\} :2^{34}$

$n = \big[ \big(2^{12}\cdot3\big)^{3} : \big(43 - 4^{4\cdot2} : 4^{4}\big)+ 2^{36}\big]:2^{34}$

$n = \big[ \big(2^{12}\big)^{3}\cdot3^{3} : \big(43 - 4^{8-4}\big)+ 2^{36}\big]:2^{34}$

$n = \big[ 2^{12\cdot3}\cdot3^{3} : \big(43 - 4^{2}\big)+ 2^{36}\big]:2^{34}$

$n = \big[ 2^{36}\cdot3^{3} : \big(43 - 16\big)+ 2^{36}\big]:2^{34}$

$n = \big(2^{36}\cdot3^{3} : 27+ 2^{36}\big):2^{34}$

$n = \big(2^{36}\cdot3^{3} : 3^{3} + 2^{36}\big):2^{34}$

$n = \big(2^{36}\cdot3^{3-3} + 2^{36}\big):2^{34}$

$n = \big(2^{36}\cdot3^{0} + 2^{36}\big):2^{34}$

$n = \big(2^{36}\cdot 1 + 2^{36}\big):2^{34}$

$n = 2^{36} \cdot\big(2^{36-36} + 2^{36-36}\big):2^{34}$

$n = 2^{36} \cdot\big(2^{0} + 2^{0}\big):2^{34}$

$n = 2^{36} \cdot\big(1 + 1\big):2^{34}$

$n = 2^{36} \cdot 2 :2^{34}$

$n = 2^{36+1-34}$

$\red{\large\boxed{~n = 2^{3}~\implies cub~ perfect}}$

$==pav38==$