B=2019.(2020-164)-2020.(2019-164)
làm hộ mik với :33
K
Khách
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Những câu hỏi liên quan
1 tháng 12 2023
\(A=\dfrac{2020^{2018}-1}{2020^{2019}+2019}\)
\(B=\dfrac{2020^{2019}+1}{2020^{2020}+2019}\)
Ta có :
\(A-B=\dfrac{2020^{2018}-1}{2020^{2019}+2019}-\dfrac{2020^{2019}+1}{2020^{2020}+2019}\)
\(\Rightarrow A-B=\dfrac{\left(2020^{2018}-1\right)\left(2020^{2020}+2019\right)-\left(2020^{2019}+2019\right)\left(2020^{2019}+1\right)}{\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)}\)
\(\Rightarrow A-B=\dfrac{2020^{4038}+2019.2020^{2018}-2020^{2020}-2019-2020^{4038}-2020^{2019}-2019.2020^{2018}-2029}{\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)}\)
\(\Rightarrow A-B=\dfrac{-\left(2020^{2020}+2020^{2019}+2.2019\right)}{\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)}\)
mà \(\left\{{}\begin{matrix}-\left(2020^{2020}+2020^{2019}+2.2019\right)< 0\\\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)>0\end{matrix}\right.\)
\(\Rightarrow A-B< 0\)
\(\Rightarrow A< B\)
Vậy ta được \(A< B\)
K
1
28 tháng 6 2021
Ta có: \(A=\left(2020^{2019}+2019^{2019}\right)^{2020}\)
\(=\left(2019^{2019}+2020^{2019}\right)^{2019}\cdot\left(2019^{2019}+2020^{2019}\right)\)
\(\Leftrightarrow\dfrac{A}{B}=\dfrac{\left(2019^{2019}+2020^{2019}\right)^{2019}\cdot\left(2019^{2019}+2020^{2019}\right)}{\left(2020^{2020}+2019^{2020}\right)^{2019}}\)
\(\Leftrightarrow\dfrac{A}{B}=\dfrac{2019^{2019}+2020^{2019}}{2019+2020}>1\)
\(\Leftrightarrow A>B\)
DQ
0
NT
3
DD
Đoàn Đức Hà
Giáo viên
26 tháng 5 2021
\(m=\frac{2019}{2020}+\frac{2020}{2021}+\frac{2021}{2019}=1-\frac{1}{2020}+1-\frac{1}{2021}+1+\frac{2}{2019}\)
\(=3+\left(\frac{1}{2019}-\frac{1}{2020}\right)+\left(\frac{1}{2019}-\frac{1}{2021}\right)>3+0+0=3\)
NN
0
\(B=2019\cdot\left(2020-164\right)-2020\left(2019-164\right)\)
\(=2019\cdot2020-2019\cdot164-2020\cdot2019+2020\cdot164\)
\(=2020\cdot164-2019\cdot164=164\)