So sánh A và B: A=$\frac{2^{2020}+1}{2^{2022}+1}$ B=$\frac{2^{2019}+1}{2^{2021}+1}$ 30/08/2021 Bởi Arya So sánh A và B: A=$\frac{2^{2020}+1}{2^{2022}+1}$ B=$\frac{2^{2019}+1}{2^{2021}+1}$
Đáp án: B>A Giải thích các bước giải: 2020A=20202022+2020/20202022+1=1+2019/20202022+1 2020B=20202021+2020=20202021+1=1+2019/20202021+1 > 1+2019/20202022+1 -> 2020B > 2020 A -> B > A Bình luận
Đáp án+Giải thích các bước giải: $A=\dfrac{2^{2020}+1}{2^{2022}+1}\\=>2^2A=\dfrac{2^{2022}+4}{2^{2022}+1}=1+\dfrac3{2^{2022}+1}$ $B=\dfrac{2^{2019}+1}{2^{2021}+1}\\=>2^2B=\dfrac{2^{2021}+4}{2^{2021}+1}=1+\dfrac3{2^{2021}+1}$ $Vì,2^{2022}+1>2^{2021}+1\\=>2^2A<2^2B\\=>A<B$ Bình luận
Đáp án:
B>A
Giải thích các bước giải:
2020A=20202022+2020/20202022+1=1+2019/20202022+1
2020B=20202021+2020=20202021+1=1+2019/20202021+1 > 1+2019/20202022+1
-> 2020B > 2020 A -> B > A
Đáp án+Giải thích các bước giải:
$A=\dfrac{2^{2020}+1}{2^{2022}+1}\\=>2^2A=\dfrac{2^{2022}+4}{2^{2022}+1}=1+\dfrac3{2^{2022}+1}$
$B=\dfrac{2^{2019}+1}{2^{2021}+1}\\=>2^2B=\dfrac{2^{2021}+4}{2^{2021}+1}=1+\dfrac3{2^{2021}+1}$
$Vì,2^{2022}+1>2^{2021}+1\\=>2^2A<2^2B\\=>A<B$