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Ta có:
\(A=\frac{2021^{2021}+1}{2021^{2022}+1}\Leftrightarrow10A=\frac{2021^{2022}+10}{2021^{2022}+1}=1+\frac{9}{2021^{2022}+1}\)
\(B=\frac{2021^{2022}-1}{2021^{2023}-1}\Leftrightarrow10B=\frac{2021^{2023}-10}{2021^{2023}-1}=1-\frac{9}{2021^{2023}-1}\)
Hay ta đang so sánh: \(\frac{9}{2021^{2022}};\frac{9}{2021^{2023}}\)
Mà \(\frac{9}{2021^{2022}}>\frac{9}{2021^{2023}}\)nên \(\frac{2021^{2021}+1}{2021^{2022}+1}>\frac{2021^{2022}-1}{2021^{2023}-1}\)hay\(A>B\)
Vậy \(A>B\)
\(=2021\cdot2\cdot\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)=4042\cdot\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)=0\)
Bài 1:
A = 1996 x 1997 x 1998 x 1999 + 2021 x 2022 x 2023 x 2024
A = (1996 x 1997) x (1998 x 1999) + (2021 x 2022) x (2023 x 2024)
A = \(\overline{..2}\) x \(\overline{..2}\) + \(\overline{..2}\) x \(\overline{..2}\)
A = \(\overline{..4}\) + \(\overline{..4}\)
A = \(\overline{..8}\)
ko ghi lại đề
ta thấy : 2019 - 1 = 2018
2020 - 2 = 2018
2021 - 3 = 2018
2022 - 4 = 2018
=> x = 2018
thử lại :
2018+1/2019 + 2018+2/2020 = 2018+3/2021 + 2018+4/2022
= 1 + 1 = 1 + 1
2 = 2
Nhỏ hơn
Ta có 2020/2021 <1
2021/2022 <1
2022/2023 <1
2023/2024 <1
Suy ra A=(2021/2021+2021/2022 +2022/2023 +2023/2024) < (1+1+1+1)= 4
Vậy A <4
Chúc bạn học tốt
\(\dfrac{2020}{2021}< 1\)
\(\dfrac{2021}{2022}< 1\)
\(\dfrac{2021}{2022}< 1\)
\(\dfrac{2023}{2024}< 1\)
Do đó: A<4
\(\frac{2011.2013+2014}{2012.2012+2013}=\frac{2011.2013+2013+1}{2012.2012+2012+1}=\frac{2013.\left(2011+1\right)+1}{2012.\left(2012+1\right)+1}=\frac{2013.2012+1}{2012.2013+1}=1\)
Vậy \(\frac{2011.2013+2014}{2012.2012+2013}=1\)
(Dấu . là nhân nha bạn)
1.
\(\dfrac{998+999\times1000}{999\times1001-1}=\dfrac{998+999\times1000}{999\times1000+999-1}=\dfrac{998+999\times1000}{999\times1000+998}=1\)
2.
\(\dfrac{2011\times2022+2023\times19+2011}{2021\times2022-2022-2020}\)
\(=\dfrac{2011\times\left(2022+1\right)+2023\times19}{\left(2021-1\right)\times2022-2020}\)
\(=\dfrac{2011\times2023+2023\times19}{2020\times2022-2020}\)
\(=\dfrac{\left(2011+19\right)\times2023}{2020\times\left(2022-1\right)}\)
\(=\dfrac{2030\times2023}{2020\times2021}\)
\(=\dfrac{410669}{408242}\)