So sánh
A. √2021 - √2020 và √2020 - √2019
B. √2019×2021 và 2020
C. √2019 + √2021 và 2√2020
K
Khách
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Những câu hỏi liên quan
SN
1
17 tháng 4 2021
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!
17 tháng 4 2021
Ta có : \(\dfrac{2019}{2020}>\dfrac{2019}{2020+2021}\)
\(\dfrac{2020}{2021}>\dfrac{2020}{2020+2021}\)
\(\Rightarrow\dfrac{2019}{2020}+\dfrac{2020}{2021}>\dfrac{2019+2020}{2020+2021}\)
\(\Rightarrow M>N\)
PN
0
TT
1
NK
1
24 tháng 5 2020
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.
9 tháng 8 2019
bài 1:
ssh của A là:
(151-3):2+1=75
A=(151+3)x75:2=5775
đáp số: 5775
XO
30 tháng 7 2020
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
PN
1
a) Ta có: \(\sqrt{2021}-\sqrt{2020}\)
\(=\frac{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}{\sqrt{2021}+\sqrt{2020}}\)
\(=\frac{1}{\sqrt{2020}+\sqrt{2021}}\)
Ta có: \(\sqrt{2020}-\sqrt{2019}\)
\(=\frac{\left(\sqrt{2020}-\sqrt{2019}\right)\left(\sqrt{2020}+\sqrt{2019}\right)}{\sqrt{2020}+\sqrt{2019}}\)
\(=\frac{1}{\sqrt{2019}+\sqrt{2020}}\)
Ta có: \(\sqrt{2020}+\sqrt{2021}>\sqrt{2019}+\sqrt{2020}\)
\(\Leftrightarrow\frac{1}{\sqrt{2020}+\sqrt{2021}}< \frac{1}{\sqrt{2019}+\sqrt{2020}}\)
hay \(\sqrt{2021}-\sqrt{2020}< \sqrt{2020}-\sqrt{2019}\)
b) Ta có: \(\sqrt{2019\cdot2021}\)
\(=\sqrt{\left(2020-1\right)\left(2020+1\right)}\)
\(=\sqrt{2020^2-1}\)
Ta có: \(2020=\sqrt{2020^2}\)
Ta có: \(2020^2-1< 2020^2\)
nên \(\sqrt{2020^2-1}< \sqrt{2020^2}\)
\(\Leftrightarrow\sqrt{2019\cdot2021}< 2020\)
c) Ta có: \(\left(\sqrt{2019}+\sqrt{2021}\right)^2\)
\(=2019+2021+2\cdot\sqrt{2019\cdot2021}\)
\(=4040+2\sqrt{2019\cdot2021}\)
\(=4040+2\cdot\sqrt{2020^2-1}\)
Ta có: \(\left(2\sqrt{2020}\right)^2\)
\(=4\cdot2020\)
\(=4040+2\cdot2020\)
\(=4040+2\cdot\sqrt{2020^2}\)
Ta có: \(2020^2-1< 2020^2\)
\(\Leftrightarrow\sqrt{2020^2-1}< \sqrt{2020^2}\)
\(\Leftrightarrow2\cdot\sqrt{2020^2-1}< 2\cdot\sqrt{2020^2}\)
\(\Leftrightarrow4040+2\cdot\sqrt{2020^2-1}< 4040+2\cdot\sqrt{2020^2}\)
\(\Leftrightarrow\left(\sqrt{2019}+\sqrt{2021}\right)^2< \left(2\sqrt{2020}\right)^2\)
\(\Leftrightarrow\sqrt{2019}+\sqrt{2021}< 2\sqrt{2020}\)