Question

Rezolva-ti exercițiul! Dau coroana!​

Answer 1

$a = {\Big ( (-n)^0 + 8^{4n+1}:16^{3n} + 6\cdot 27^{10}:(-3)^{28}\Big )}^{63} \\\\={\Big ( 1 + 8^{4n}\cdot 8^{1}:16^{3n} + 2\cdot 3\cdot {3^3}^{10}:{3}^{28}\Big )}^{63} \\\\= {\Big ( 1 + 16^{2n}\cdot 8^{1}:16^{3n} + 2\cdot 3\cdot {3}^{30}:{3}^{28}\Big )}^{63} \\\\={\Big ( 1 + 8:16^{n} + 2\cdot {3}^{3}\Big )}^{63} \\\\={\Big ( 1 + 8:16^n + 2\cdot {3}^{3}\Big )}^{63} \\\\={\Big ( 1 + 8:(16^n) + 2\cdot {3}^{3}\Big )}^{63}\\\\ = {\Big ( 1 + 8:(16^n) + 54\Big )}^{63} \\\\={\Big ( 55 + 8:(16^n)\Big )}^{63}$

$b={\Big( 2^{2^5} : 2^{5^2} - 3^{2^2} + 2^{2^2} \Big)}^{62}\\\\={\Big( 2^{2^5 - 5^2} - 3^{4} + 2^{4} \Big)}^{62}\\\\ = {\Big( 2^{32 - 25} - 81 + 16 \Big)}^{62}\\\\={\Big( 2^{7} - 81 + 16 \Big)}^{62} \\\\= {\Big( 128 - 65 \Big)}^{62}\\\\={\Big( 63 \Big)}^{62} = 63^{62}$

$\frac{a}{b} = \frac{ {\Big ( 1 + 8:(16^n) + 54\Big )}^{63} }{ 63^{62} } = \Big ( 1 + 8:(16^n) + 54\Big )\cdot \frac{ {\Big ( 1 + 8:(16^n) + 54\Big )}^{62} }{63^{62}} = \Big ( 1 + 8:(16^n) + 54\Big )\cdot {\Bigg(\frac{\Big ( 1 + 8:(16^n) + 54\Big )}{63}\Bigg)}^{62} = \Big ( 55 + 8:(16^n)\Big )\cdot {\Bigg(\frac{\Big ( 55 + 8:(16^n) \Big )}{63}\Bigg)}^{62}$