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Ta có:
2019.2021=2019.(2020+1)=2019.2020+2019 (1)
Lại có:
2020.2020=(2019+1).2020=2019.2020+2020 (2)
Vì 2019.2020=2019.2020 mà 2019<2020
=>(1)<(2)
=>.....
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Ta có: \(\frac{a}{b}+\frac{b}{a}\le2\)
Dấu bằng xảy ra khi : a=b
=>\(\frac{2021}{2019}+\frac{2019}{2021}< 2\)
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Giải:
Ta có: N=2019+2020/2020+2021
=>N=2019/2020+2021 + 2020/2020+2021
Vì 2019/2020 > 2019/2020+2021 ; 2020/2021 > 2020/2020+2021
=>M>N
Vậy ...
Chúc bạn học tốt!
Ta có : \(\dfrac{2019}{2020}>\dfrac{2019}{2020+2021}\)
\(\dfrac{2020}{2021}>\dfrac{2020}{2020+2021}\)
\(\Rightarrow\dfrac{2019}{2020}+\dfrac{2020}{2021}>\dfrac{2019+2020}{2020+2021}\)
\(\Rightarrow M>N\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$A=1-\frac{1}{2019}+1-\frac{1}{2020}+1-\frac{1}{2021}+1+\frac{3}{2018}$
$=4+(\frac{1}{2018}-\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2020}+\frac{1}{2018}-\frac{1}{2021})$
$> 4+0+0+0+0=4$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{2021}{2019}và\dfrac{2023}{2021}\)
\(\Rightarrow\dfrac{2021}{2019}-\dfrac{2}{2019}=\dfrac{2023}{2021}-\dfrac{2}{2021}\left(=1\right)\)
\(\Rightarrow\dfrac{2}{2019}>\dfrac{2}{2021}\Rightarrow\dfrac{2021}{2019}< \dfrac{2023}{2021}\)
Chứng minh bđt phụ nếu a>b \(\Rightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\left(vớim\in N^{\circledast}\right)\Rightarrow a\left(b+m\right)>b\left(a+m\right)\Rightarrow ab+am>ab+bm\Rightarrow am>bm\Rightarrow a>b\) \(\Rightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\left(1\right)\)
Áp dụng bđt (1) có :
\(2021>2019\Rightarrow\dfrac{2021}{2019}>\dfrac{2021+2}{2019+2}=\dfrac{2023}{2021}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{2019}{2021}< 1\)
\(\frac{2021}{2019}>1\)
\(\Rightarrow\frac{2019}{2021}+\frac{2021}{2019}< 2\)