Question

Exercitiul 7 a si b va rog .

Answer 1

Explicație pas cu pas:

7.a)

${2}^{100} + {2}^{101} + {2}^{102} = {2}^{100} \cdot (1 + 2 + {2}^{2}) = {2}^{2 \cdot 50} \cdot (1 + 2 + 4) = {( {2}^{2} )}^{50} \cdot 7 = 7 \cdot {4}^{50}$

${5}^{52} - {5}^{51} - {5}^{50} = {5}^{50} \cdot ( {5}^{2} - 5 - 1) = {5}^{50} \cdot (25 - 5 - 1) = 19 \cdot {5}^{50}$

$7 < 19 \ \ si \ \ {4}^{50} < {5}^{50} \implies \bf 7 \cdot {4}^{50} < 19 \cdot {5}^{50}$

=>

$\bf {2}^{100} + {2}^{101} + {2}^{102} < {5}^{52} - {5}^{51} - {5}^{50}$

b)

${2}^{77} - {2}^{76} = {2}^{76} \cdot (2 - 1) = {2}^{76} = 2 \cdot {2}^{5 \cdot15} = 2 \cdot {32}^{15}$

${3}^{46} - {27}^{15} = {3}^{46} - {( {3}^{3} )}^{15} = {3}^{46} - {3}^{45} = {3}^{45} \cdot (3 - 1) = 2 \cdot {3}^{45} = 2 \cdot {3}^{3 \cdot 15} = 2 \cdot {27}^{15}$

$2 \cdot {32}^{15} > 2 \cdot {27}^{15}$

=>

$\bf {2}^{77} - {2}^{76} > {3}^{46} - {27}^{15}$