Cho A=1+2020+2020^2+...+2020^98
So sanh A voi 2020.2020^98-2
Giup minh nhe!
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Ta có: \(\frac{A}{2020}=\frac{2020^{100}-1}{2020^{100}+2020}=\frac{2020^{100}+2020-2021}{2020^{100}+2020}=1-\frac{2021}{2020^{100}+2020}\)
\(\frac{B}{2020}=\frac{2020^{101}-1}{2020^{101}+2020}=\frac{2020^{101}+2020-2021}{2020^{101}+2020}=1-\frac{2021}{2020^{101}+2020}\)
Ta có: \(2020^{100}+2020<2020^{101}+2020\)
=>\(\frac{2021}{2020^{100}+2020}>\frac{2021}{2020^{101}+2020}\)
=>\(-\frac{2021}{2020^{100}+2020}<-\frac{2021}{2020^{101}+2020}\)
=>\(-\frac{2021}{2020^{100}+2020}+1<-\frac{2021}{2020^{101}+2020}+1\)
=>\(\frac{A}{2020}<\frac{B}{2020}\)
=>A<B
Ta có:
\(\frac{2017}{2019}=1-\frac{2}{2019}\)
\(\frac{2018}{2020}=1-\frac{2}{2020}\)
Vì \(\frac{2}{2019}>\frac{2}{2020}\)
=> \(1-\frac{2}{2019}>1-\frac{2}{2020}\)
=> \(\frac{2017}{2019}>\frac{2018}{2020}\)
\(10A=\dfrac{10^{2021}+1+9}{10^{2021}+1}=1+\dfrac{9}{10^{2021}+1}\)
\(10B=\dfrac{10^{2022}+1+9}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)
mà \(10^{2021}+1< 10^{2022}+1\)
nên A>B
Minh tinh ra duoc la:
a=2020^99-2/2019
2020^99-1/2019 2020^99-2
2020^99-1 2020^99*2019-2*2019
2*2019-1 2020^99(2019-1)
2*2019-2 2020^99*2018
Gio phai lam sao tiep moi nguoi?