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\(\dfrac{1}{2020}-\dfrac{1}{2021}=\dfrac{2021}{2020.2021}-\dfrac{2020}{2020.2021}=\dfrac{2021-2020}{2020.2021}=\dfrac{1}{2020.2021}\)
\(\dfrac{1}{2020\cdot2021}=\dfrac{2021-2020}{2020\cdot2021}=\dfrac{1}{2020}-\dfrac{1}{2021}\)(đpcm)
2019 . 2021 - 2020. 2020
= 2019(2020+1) - 2020( 2019+1)
= 2019 .2020 + 1.2019 - 2020.2019 + 1.2020
= 2019 -2020
= -1
\(2019.2021-2020.2020\)
\(=2019\left(2020+1\right)-2020\left(2019+1\right)\)
\(=2019.2020+2019-2020.2019+2020\)
\(=2019-2020\)
\(=-1\)
À câu này giải theo cách này e nhé!
E tham khảo nha:(-- được xem là +)
=>(2020+27)+(2020-73)-129
=2047+1947-129=3865
\(t=\frac{2009.2010+2000}{2011.2010-2020}\)
\(t=\frac{2009.2010+2000}{\left(2009+2\right).2010-2020}\)
\(t=\frac{2009.2010+2000}{2009.2010+2.2010-2020}\)
\(t=\frac{2009.2010+2000}{2009.2010+2000}=1\)
\(q=\frac{\text{2014.2015+2010}}{2016.2015-2020}\)
\(q=\frac{\text{2014.2015+2010}}{\left(2014+2\right).2015-2020}\)
\(q=\frac{\text{2014.2015+2010}}{2014.2015+2.2015-2020}\)
\(q=\frac{\text{2014.2015+2010}}{\text{2014.2015+2010}}=1\)
2020/2021 < 1 < 2021/2020
Suy ra 2020/2021 < 2021/2020
mk sai đề một tí
A=2019 mũ 2020 + 1
trên 2019 mũ 2020 - 3
B=2019 mũ 2020 -1
trên 2019mũ 2020 - 5
so sánh A và B
Ta có: \(A=\left(2020^{2019}+2019^{2019}\right)^{2020}\)
\(=\left(2019^{2019}+2020^{2019}\right)^{2019}\cdot\left(2019^{2019}+2020^{2019}\right)\)
\(\Leftrightarrow\dfrac{A}{B}=\dfrac{\left(2019^{2019}+2020^{2019}\right)^{2019}\cdot\left(2019^{2019}+2020^{2019}\right)}{\left(2020^{2020}+2019^{2020}\right)^{2019}}\)
\(\Leftrightarrow\dfrac{A}{B}=\dfrac{2019^{2019}+2020^{2019}}{2019+2020}>1\)
\(\Leftrightarrow A>B\)