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Ta có: \(A=\frac{2020}{2021}+\frac{2021}{2022}\)
\(\Rightarrow A=\frac{2021}{2021}-\frac{1}{2021}+\frac{2022}{2022}-\frac{1}{2022}\)
\(\Rightarrow A=1-\frac{1}{2021}+1-\frac{1}{2022}\)
\(\Rightarrow A=1+1-\frac{1}{2021}-\frac{1}{2022}\)
\(\Rightarrow A=2-\frac{1}{2021}-\frac{1}{2022}\)
\(\Rightarrow A=2-\frac{1}{2021\cdot2022}\)
\(B=\frac{2020+2021}{2021+2022}\)
\(\Rightarrow B=\frac{2021+2022}{2021+2022}-\frac{2}{2021+2022}\)
\(\Rightarrow B=1-\frac{2}{2021+2022}\)
\(\Rightarrow B=1-\frac{2}{4043}\)
Vậy ta sẽ so sánh:
\(1-\frac{1}{2021\cdot2022};\frac{2}{4043}\)
Vì \(2021\cdot2022>4043\)nên \(\frac{1}{2021\cdot2022}< \frac{2}{4043}\)vậy \(1-\frac{1}{2021\cdot2022}>\frac{2}{4043}\)
\(\Rightarrow\frac{2020}{2021}+\frac{2021}{2022}>\frac{2020+2021}{2021+2022}\)
\(\Rightarrow A>B\)
Ta có : A = \(\frac{10^{2020}+1}{10^{2021}+1}\)
=> 10A = \(\frac{10^{2021}+10}{10^{2021}+1}=1+\frac{9}{10^{2021}+1}\)
Lại có : \(B=\frac{10^{2021}+1}{10^{2022}+1}\)
=> \(10B=\frac{10^{2022}+10}{10^{2022}+1}=1+\frac{9}{10^{2022}+1}\)
Vì \(\frac{9}{10^{2022}+1}< \frac{9}{10^{2021}+1}\)
=> \(1+\frac{9}{10^{2022}+1}< 1+\frac{9}{10^{2022}+1}\)
=> 10B < 10A
=> B < A
b) Ta có : \(\frac{2019}{2020+2021}< \frac{2019}{2020}\)
Lại có : \(\frac{2020}{2020+2021}< \frac{2020}{2021}\)
=> \(\frac{2019}{2020+2021}+\frac{2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)
=> \(\frac{2019+2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)
=> B < A
Ta có: \(\frac{2019}{2020}>\frac{2019}{2020+2021};\frac{2020}{2021}>\frac{2020}{2020+2021}\)
=> \(\frac{2019}{2020}+\frac{2020}{2021}>\frac{2019}{2020+2021}+\frac{2020}{2020+2021}=\frac{2019+2020}{2020+2021}\)
=> A > B.
N =2019+2020/2020+2021
=2019/2020+2021 + 2020/2020+2021
Ta có:
2019/2020>2019/2020+2021
2020/2021 > 2020/2020+2021
=>M>N
Ta có : \(\frac{2019}{2020}=1-\frac{1}{2020}\)
\(\frac{2020}{2021}=1-\frac{1}{2021}\)
Vì \(\frac{1}{2020}>\frac{1}{2021}\) nên \(1-\frac{1}{2020}< 1-\frac{1}{2021}\)
\(\Rightarrow\frac{2019}{2020}< \frac{2020}{2021}\)
Ta có : \(\frac{672}{2017}< \frac{673}{2017}< \frac{673}{2020}\)
\(\frac{\Rightarrow672}{2017}< \frac{673}{2020}\)
1.So sánh phân số: \(\frac{2019}{2020}\) và \(\frac{2020}{2021}\)
Ta có : \(\frac{2019}{2020}\) + \(\frac{1}{2020}\) = \(\frac{2020}{2020}\) = 1
\(\frac{2020}{2021}\) + \(\frac{1}{2021}\) = \(\frac{2021}{2021}\) = 1
Mà \(\frac{1}{2020}\) > \(\frac{1}{2021}\) nên \(\frac{2019}{2020}\) < \(\frac{2020}{2021}\)
Mình chỉ biết mỗi câu này thôi, mình chắc chắn với bạn là câu này đúng không sai đâu
~ Học tốt ~
a) Ta có A = \(\frac{2^{2018}+1}{2^{2019}+1}\)
=> 2A = \(\frac{2^{2019}+2}{2^{2019}+1}=1+\frac{1}{2^{2019}+1}\)
Lại có B = \(\frac{2^{2017}+1}{2^{2018}+1}\)
=> 2B = \(\frac{2^{2018}+2}{2^{2018}+1}=\frac{2^{2018}+1+1}{2^{2018}+1}=1+\frac{1}{2^{2018}+1}\)
Vì \(\frac{1}{2^{2018}+1}>\frac{1}{2^{2019}+1}\Rightarrow1+\frac{1}{2^{2018}+1}>1+\frac{1}{2^{2019}+1}\Rightarrow2B>2A\Rightarrow B>A\)
Ta có :
\(N=\frac{2018+2019+2020}{2019+2020+2021}\)
\(=\frac{2018}{2019+2020+2021}+\frac{2019}{2019+2020+2021}+\frac{2020}{2019+2020+2021}\)
Mà \(\frac{2018}{2019}>\frac{2018}{2019+2020+2021}\)
\(\frac{2019}{2020}>\frac{2019}{2019+2020+2021}\)
\(\frac{2020}{2021}>\frac{2020}{2019+2020+2021}\)
\(\Leftrightarrow M>N\)
Trả lời:
Ta có:
\(\frac{2018}{2019}>\frac{2018}{2019+2020+2021}\)
\(\frac{2019}{2020}>\frac{2019}{2019+2020+2021}\)
\(\frac{2020}{2021}>\frac{2020}{2019+2020+2021}\)
\(\Rightarrow\frac{2018}{2019}+\frac{2019}{2020}+\frac{2020}{2021}>\frac{2018+2019+2020}{2019+2020+2021}\)
hay \(M>N\)
Vậy \(M>N\)
\(A=\frac{2019^{2020}+1}{2019^{2021}+1}\)và \(B=\frac{2019^{2018}+1}{2019^{2019}+1}\)
Xét \(A=\frac{2019^{2020}+1}{2019^{2021}+1}\Rightarrow2019A=\frac{2019^{2021}+2019}{2019^{2021}+1}=1+\frac{2019}{2019^{2021}+1}\)
Xét \(B=\frac{2019^{2018}+1}{2019^{2019}+1}\Rightarrow2019B=\frac{2019^{2019}+2019}{2019^{2019}+1}=1+\frac{2018}{2019^{2019}+1}\)
Vì \(1+\frac{2018}{2019^{2021}+1}< 1+\frac{2018}{2019^{2019}+1}\Rightarrow\frac{2019^{2020}+1}{2019^{2021}+1}< \frac{2018^{2019}+1}{2019^{2019}+1}\)
\(\Rightarrow A< B\)
Ta có:
\(A=\frac{2019^{2020}+1}{2019^{2021}+1}\)
\(\Rightarrow2019A=\frac{2019^{2021}+2019}{2019^{2021}+1}\)
\(\Rightarrow2019A=1+\frac{2019}{2019^{2021}+1}\)
\(\Rightarrow A=1+\frac{2019}{2019^{2021}+1}:2019\)
Ta lại có:
\(B=\frac{2019^{2018}+1}{2019^{2019}+1}\)
\(\Rightarrow2019B=\frac{2019^{2019}+2019}{2019^{2019}+1}\)
\(\Rightarrow2019B=1+\frac{2019}{2019^{2019}+1}\)
\(\Rightarrow B=1+\frac{2019}{2019^{2019}+1}:2019\)
Do \(2019^{2021}+1>2019^{2019}+1\)
\(\Rightarrow\frac{2019}{2019^{2021}+1}< \frac{2019}{2019^{2019}+1}\)
\(\Rightarrow1+\frac{2019}{2019^{2021}+1}:2019< 1+\frac{2019}{2019^{2019}+1}:2019\)
\(\Rightarrow A< B\)
Vậy \(A< B.\)
2022/2021 = 1 + 1/2021
2021/2020 = 1 + 1/2020
Do 2021 > 2020 nên 1/2021 < 1/2020
⇒ 1 + 1/2021 < 1 + 1/2020
⇒ 2022/2021 < 2021/2020
Có:
\(\dfrac{2022}{2021}-1=\dfrac{2022-2021}{2021}=\dfrac{1}{2021}\)
\(\dfrac{2021}{2020}-1=\dfrac{2021-2020}{2020}=\dfrac{1}{2020}\)
Ta thấy: \(2021>2020\)\(\Rightarrow\dfrac{1}{2021}< \dfrac{1}{2020}\)
hay \(\dfrac{2022}{2021}-1< \dfrac{2021}{2020}-1\Rightarrow\dfrac{2022}{2021}< \dfrac{2021}{2020}\)