32019 : (32020 - 24.32017)
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`#3107`
\(\left(3^{2021}+3^{2020}\right)\div3^{2020}\\ =3^{2021}\div3^{2020}+3^{2020}\div3^{2020}\\ =3^{2021-2020}+3^{2020-2020}\\ =3+1=4\)
`@` `\text {Ans}`
`\downarrow`
`a)`
Ta có: `2020` là lũy thừa bậc chẵn
`=>`\(\left(-3\right)^{2020}=3^{2020}\)
`M = `\(3^{2020}-3^{2020}=0\)
`=> 0 = 0`
`=> M = N`
`b)`
`M =`\(\left(-3\right)^{2021}+3^{2020}\)
`=`\(3^{2020}-3^{2021}\)
Vì \(3^{2021}>3^{2020}\)
`=>`\(3^{2020}-3^{2021}< 0\)
`N = [ (-3)]^0`
`= (-3)^0`
`= 1`
Vì `1 > 0`
`=> M < N.`
`@` `\text {Duynamlvhg}`
a: M=3^2020-3^2020=0
b: M=-3^2021+3^2020=-3^2020(3-1)=-3^2020*2<0
N=[(-3)]^0=1
=>M<N
\(\Rightarrow12x-33=3\\ \Rightarrow12x=36\\ \Rightarrow x=3\)
A=32019+1+3+32+33+...+32018
⇒A=1+3+32+...+32018+32019
⇒3A=3×(1+3+3^2+3^3+....+3^2019)
3A=3+3^2+3^3+....+3^2020
3A-A=(3+3^2+3^3+....+3^2020) -(1+3+3^2+....+3^2019)
2A= 3^2020-1
⇒ A =( 3^2020-1):2
A=32019+1+3+32+33+...+32018
⇒A=1+3+32+...+32018+32019
⇒3A=3×(1+3+3^2+3^3+....+3^2019)
⇒3A=3+3^2+3^3+....+3^2020
⇒3A-A=(3+3^2+3^3+....+3^2020) -(1+3+3^2+....+3^2019)
⇒2A= 3^2020-1
⇒ A =( 3^2020-1):2
Ghi lại đề: \(A=3+3^2+...+3^{2020}\)
\(\Rightarrow A=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2017}+3^{2018}+3^{2019}+3^{2020}\right)\\ A=3\left(1+3+3^2+3^3\right)+...+3^{2017}\left(1+3+3^2+3^3\right)\\ A=\left(1+3+3^2+3^3\right)\left(3+...+3^{2017}\right)\\ A=40\left(3+...+3^{2017}\right)⋮10\left(40⋮10\right)\)
bằng 3
Tính
32019 : (32020 - 24 . 32017)
= 32019 : (32020 - 8 . 32018)
= 32019 : [32018 (32 - 8)]
= 32019 : 32018
= 3