Question

ma ajuta și pe mine cnv la ex acesta? ​

Answer 1

$a)\quad U(3^{2009}+5^{2020}+1^{2021}) = U(3^{2\cdot 1004+1}+5+1) = \\ \\ =U(9^{1004}\cdot 3+6) = U(81^{502}\cdot 3+6) = U(1\cdot 3+6) = \boxed{9}\\ \\ \\ b)\quad U(2^{2016}+6^{2020}+2019^0) = U(4^{1008}+6+1) = \\ \\ = U(16^{504}+7) = U(6+7) = \boxed{3}\\ \\ \\ c)\quad U(4^{2019}+7^{2020}+8^{2021}-9^{2022}) = \\ \\ = U(16^{1009}\cdot 4+49^{1010}+64^{1010}\cdot 8-81^{1011}) = \\ \\ =U(6\cdot 4+81^{505}+16^{505}\cdot 8 - 1) = \\ \\ = U(4+1+6\cdot 8-1) = \\ \\ =U(52)= \\ \\ = U(2)\quad (F)$

$9^{2022} = 9\cdot 9^{2021} = 9^{2021}+9^{2021}+9^{2021}+6\cdot 9^{2021} > \\ \\ >8^{2021}+7^{2020}+4^{2019} \Rightarrow \\ \\ \Rightarrow 0 > 4^{2019}+7^{2020}+8^{2021}-9^{2022}\\ \\ \\ \text{Deoarece }4^{2019}+7^{2020}+8^{2021}-9^{2022} < 0 \\ \\ \Rightarrow U(4^{2019}+7^{2020}+8^{2021}-9^{2022}) = U(10-2) = \boxed{8}$