Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{2022\times2021-1}{2020\times2022+2021}\)
\(=\dfrac{2022\times2021-1}{2021\times2022-2022+2021}\)
\(=\dfrac{2022\times2021-1}{2021\times2022-1}\)
\(=1\)
\(a,50\%+\dfrac{7}{12}-\dfrac{1}{2}\\ =\dfrac{1}{2}+\dfrac{7}{12}-\dfrac{1}{2}\\ =\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\dfrac{7}{12}\\ =\dfrac{7}{12}\\ b,2022\times67+2022\times43-2022\times10\\ =2022\times\left(67+43-10\right)\\ =2022\times100\\ =202200.\\ c,125-25:3\times12\)
\(=25\times5-25:3\times12\\ =25\times\left(5-\dfrac{1}{3}\right)\times12\\ =25\times\dfrac{14}{3}\times12\\ =1400\)
a,50%+127−21=21+127−21=(21−21)+127=127b,2022×67+2022×43−2022×10=2022×(67+43−10)=2022×100=202200.c,125−25:3×12
\(2022\times2005-2000\times2022+15\times2022-20\times2021\)
\(=2022\times\left(2005-2000+15\right)-20\times2021\)
\(=2022\times20-20\times2021\)
\(=20\times\left(2022-2021\right)\)
\(=20\times1\)
\(=20\)
a, 2022 \(\times\) 2005 - 2000 \(\times\) 2022 + 15 \(\times\) 2022 - 20 \(\times\) 2021
= (2022 \(\times\) 2005 - 2000 \(\times\) 2022 + 15 \(\times\) 2022 )- 20 \(\times\) 2021
= 2022 \(\times\) (2005 - 2000 + 15) - 20 \(\times\) 2021
= 2022 \(\times\) (5 +15) - 20 \(\times\) 2021
= 2022 \(\times\) 20 - 20 \(\times\) 2021
= 20 \(\times\) (2022 - 2021)
= 20 \(\times\) 1
= 20
A = \(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\) + \(\dfrac{1}{3\times4}\)+...+ \(\dfrac{1}{2021\times2022}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+...+ \(\dfrac{1}{2021}\) - \(\dfrac{1}{2022}\)
A = 1 - \(\dfrac{1}{2022}\)
A = \(\dfrac{2021}{2022}\)
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 \(A>B\)
\(=2021\cdot2\cdot\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)=4042\cdot\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)=0\)
=4090506-\(\dfrac{3}{2023}\)*2021+2020
=4090506-\(\dfrac{6063}{2023}\)+2020
=\(\dfrac{8279174035}{2023}\) tối giản luôn rồi nhé
a. \(\dfrac{2021+2020.2022}{2021.2022-1}\)
\(\dfrac{2021.2022-2022+2021}{2021.2022-1}=\dfrac{2021.2022-1}{2021.2022-1}=1\)
\(b.\dfrac{2022+2021.2023}{2022.2023-1}=\dfrac{2021.2023-2023+2022}{2022.2023-1}\)
\(=\dfrac{2021.2023-1}{2022.2023-1}\)
thanks các cậu nhiều