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\(=2021\cdot2\cdot\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)=4042\cdot\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)=0\)
\(...=2021.\left(1000...000-1\right)\) (21 chữ số 0)
\(...=2021.\left(10^{2021}-1\right)\)
\(A=2018\times2020+2021\) và \(B=2019\times2019+2021\)
\(A=2018\times2019+2018+2021\)
\(B=2018\times2019+2019+2021\)
Vì \(2019>2018\Rightarrow A< B\)
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 x 2021 - 2019 x 2023
= (2019 +2) x ( 2023 -2) - 2019 x 2023
= 2019 x 2023 - 2 x 2019 + 2 x 2023 - 4 - 2019 x 2023
= ( 2019 x 2023 - 2019 x 2023) + 2 x ( 2023 - 2019) - 4
= 0 + 2 x 4 - 4
= 8 - 4
= 4
2021 x 2021 - 2019 x 2023
= (2019 +2) x ( 2023 -2) - 2019 x 2023
= 2019 x 2023 - 2 x 2019 + 2 x 2023 - 4 - 2019 x 2023
= ( 2019 x 2023 - 2019 x 2023) + 2 x ( 2023 - 2019) - 4
= 0 + 2 x 4 - 4
= 8 - 4
= 4
2021x0.5+2021x1%-2021:2
=2021x0.5+2021x0.01-2021x0.5
=2021x(0.5+0.01-0.5)
=2021x 0.01
=20,21
=20,21