tính nhanh
2019 . 2021 - 2020. 2020
K
Khách
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Những câu hỏi liên quan
KS
1
17 tháng 5 2021
2020 × 2021 - 1000 - 2020 × 2019 - 1020
= 2020 × 2021 - 2020 × 2019 - 1000 - 1020
= 2020 × 2021 - 2020 × 2019 - (1000 + 1020)
= 2020 × 2021 - 2020 × 2019 - 2020
= 2020 × 2021 - 2020 × 2019 - 2020 × 1
= 2020 × (2021 - 2019 - 1)
= 2020 × 1
= 2020.
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 : \(\dfrac{2019}{2020}>\dfrac{2019}{2020+2021}\)
\(\dfrac{2020}{2021}>\dfrac{2020}{2020+2021}\)
\(\Rightarrow\dfrac{2019}{2020}+\dfrac{2020}{2021}>\dfrac{2019+2020}{2020+2021}\)
\(\Rightarrow M>N\)
VN
3
24 tháng 2 2023
(1+3+5+7+...+2019+2021)
A=1−3+5−7+......−2019+2021−2023
A=(1−3)+(5−7)+....+(2021−2023)A=(1−3)+(5−7)+....+(2021−2023)
A=−2+(−2)+....+(−2)(506)A=−2+(−2)+....+(−2)(506cặp)
a=−2.506A=−2.506
A=−1012A=−1012
9 tháng 8 2019
bài 1:
ssh của A là:
(151-3):2+1=75
A=(151+3)x75:2=5775
đáp số: 5775
11 tháng 11 2021
\(a^{2019}+b^{2019}=a^{2020}+b^{2020}\\ \Leftrightarrow a^{2020}-a^{2019}=b^{2019}-b^{2020}=0\\ \Leftrightarrow a^{2019}\left(a-1\right)=b^{2019}\left(1-b\right)\\ \Leftrightarrow\dfrac{a^{2019}}{b^{2019}}=\dfrac{1-b}{a-1}\left(1\right)\\ a^{2020}+b^{2020}=a^{2021}+b^{2021}\\ \Leftrightarrow a^{2021}-a^{2020}=b^{2020}-b^{2021}\\ \Leftrightarrow a^{2020}\left(a-1\right)=b^{2020}\left(1-b\right)\\ \Leftrightarrow\dfrac{a^{2020}}{b^{2020}}=\dfrac{1-b}{a-1}\left(2\right)\\ \left(1\right)\left(2\right)\Leftrightarrow\dfrac{a^{2019}}{b^{2019}}=\dfrac{a^{2020}}{b^{2020}}\Leftrightarrow\dfrac{a}{b}=1\Leftrightarrow a=b\\ \Leftrightarrow2a^{2019}=2a^{2020}\\ \Leftrightarrow a=1=b\\ \Leftrightarrow P=2022-\left(1+1-1\right)^{2022}=2021\)
PN
0
4 tháng 10 2020
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: \(2020^2-1< 2020^2\)
\(\Leftrightarrow\sqrt{2020^2-1}< \sqrt{2020^2}\)
\(\Leftrightarrow2\cdot\sqrt{2020^2-1}< 2\cdot\sqrt{2020^2}\)
\(\Leftrightarrow4040+2\cdot\sqrt{2020^2-1}< 4040+2\cdot\sqrt{2020^2}\)
\(\Leftrightarrow\left(\sqrt{2019}+\sqrt{2021}\right)^2< \left(2\sqrt{2020}\right)^2\)
\(\Leftrightarrow\sqrt{2019}+\sqrt{2021}< 2\sqrt{2020}\)
2019 . 2021 - 2020. 2020
= 2019(2020+1) - 2020( 2019+1)
= 2019 .2020 + 1.2019 - 2020.2019 + 1.2020
= 2019 -2020
= -1
\(2019.2021-2020.2020\)
\(=2019\left(2020+1\right)-2020\left(2019+1\right)\)
\(=2019.2020+2019-2020.2019+2020\)
\(=2019-2020\)
\(=-1\)