m=(1-1000/2018)×(1-1001/2018)×(1-1002/2018)...(1-2019/2018) chỉ giúp vs nè
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\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018\times2019}\)
\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(A=\left(\dfrac{2020}{2019}-\dfrac{1}{2019}\right)-\left(\dfrac{2019}{2018}-\dfrac{1}{2018}\right)\)
\(A=\left(\dfrac{2020-1}{2019}\right)-\left(\dfrac{2019-1}{2018}\right)\)
\(A=1-1\)
\(A=0.\)
\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018\times2019}\)
\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(A=\left(\dfrac{2020}{2019}-\dfrac{1}{2019}\right)-\left(\dfrac{2019}{2018}-\dfrac{1}{2018}\right)\)
\(A=\dfrac{2019}{2019}-\dfrac{2018}{2018}\)
\(A=1-1\)
\(A=0\)
\(A=\frac{2018^{2019}-1}{2018^{2019}+1}=\frac{2018^{2019}+1-2}{2018^{2019}+1}=\frac{2018^{2019}+1}{2018^{2019}+1}-\frac{2}{2018^{2019}+1}=1-\frac{2}{2018^{2019}+1}\)
\(B=\frac{2018^{2019}}{2018^{2019}+2}=\frac{2018^{2019}+2-2}{2018^{2019}+2}=\frac{2018^{2019}+2}{2018^{2019}+2}-\frac{2}{2018^{2019}+2}=1-\frac{2}{2018^{2019}+2}\)
Ta có: \(\frac{2}{2018^{2019}+1}>\frac{2}{2018^{2019}+2}\)
\(\Rightarrow1-\frac{2}{2018^{2019}+1}< 1-\frac{2}{2018^{2019}+2}\)
\(\Rightarrow A< B\)
Vậy .....
\(M=\frac{2018^{2018}+1}{2019^{2019}+1}\)
\(\Leftrightarrow2M=1+\frac{2017}{2018^{2019}+1}\)
\(N=\frac{2018^{2019}-2}{2018^{2020}-2}\)
\(\Leftrightarrow2N=1-\frac{4034}{2018^{2020}-2}\)
Nhận thấy : \(1+\frac{2017}{2018^{2019}+1}>1-\frac{4034}{2018^{2020}-2}\Leftrightarrow2M>2N\Leftrightarrow M>N\)
Từ đề bài, ta suy ra:
So sánh hai biểu thức
\(M=\left(2018^{2018}+1\right)\cdot\left(2018^{2020}-2\right)\)(1)
\(N=\left(2018^{2019}-2\right)\cdot\left(2018^{2019}+1\right)\)(2)
Xét biểu thức M và N, ta suy ra:
\(M=\left(2018^{2019}-2017\right)\cdot\left(2019^{2019}+2016\right)\)
\(N=\left(2018^{2019}-2017\right)\cdot\left(2018^{2018}-2016\right)\)
Nhận thấy (20192019+2016)>(20182018-2016) nên M>N
Vậy M>N.
P/s:Mình đây không phải top 10 tuần nên bài có thể sai sót, mong bạn tham khảo:)))
2018 A = 2018 - 2018^2 + 2018^3 +...- 2018^2018 + 2018^2019
=> A + 2018 A = 1 +2018^2019
=> 2019 A = 1 + 2018^2019
=> 2019 A - 1 = 2018^2019
=> 2019 A -1 là 1 lũy thừa của 2018
\(\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2019}x2018\)
\(=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{2018}{2019}=2-\dfrac{2019}{2018}=\dfrac{2017}{2018}\)