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Ta có :
\(A=\dfrac{2019\times2020}{2019\times2020+1}=\dfrac{2019\times2020+1-1}{2019\times2020+1}=1-\dfrac{1}{2019\times2020+1}\)
Suy ra A < 1 (1)
Lại có \(B=\dfrac{2020}{2019}=\dfrac{2019+1}{2019}=\dfrac{2019}{2019}+\dfrac{1}{2019}=1+\dfrac{1}{2019}\)
Suy ra B > 1 (2)
Từ (1) và (2) ta có : A < 1 < B
=> A < B
Vậy A < B
? x ? = ?
=?
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=...................
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Ta có \(\frac{2018\times2019+4036}{2019\times2020-2}\)
\(=\frac{\left(2020-2\right)\times2019}{2019\times2020-2}\)
\(=\frac{2020\times2019-2\times2019+4036}{2019\times2020-2}\)
\(=\frac{2020\times2019-4038+4036}{2019\times2020-2}\)
\(=\frac{2020\times2019-2}{2019\times2020-2}\)
\(=1\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}>1\)
\(C=\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+...+\dfrac{2}{2019\times2020}\)
\(=2\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{2019\times2020}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\right)\)
\(=2\left(1-\dfrac{1}{2020}\right)=2.\dfrac{2019}{2020}=\dfrac{2019}{1010}\)
ko ghi lại đề
ta thấy : 2019 - 1 = 2018
2020 - 2 = 2018
2021 - 3 = 2018
2022 - 4 = 2018
=> x = 2018
thử lại :
2018+1/2019 + 2018+2/2020 = 2018+3/2021 + 2018+4/2022
= 1 + 1 = 1 + 1
2 = 2
Dấu ''\(x\)'' là dấu nhân chăng ?
\(A=\frac{2019x2020}{2019x2020+1}\)và \(B=\frac{2020}{2021}\)
Bài ra ta có :
Xét \(A=\frac{2019x2020}{2019x\left(2020+1\right)}=\frac{2020}{2020+1}=\frac{2020}{2021}\)
Vì \(\frac{2020}{2021}=\frac{2020}{2021}\)
Suy ra A = B theo (ĐPCM)