KT
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Khách
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14 tháng 5 2019
B= 1/1.2+1/2.3+...+1/2019.2020
B=1/1-1/2+1/2-1/3+...+1/2019-1/2020
B=1-1/2020=2020/2020-1/2020=2019/2020
14 tháng 5 2019
\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1.\)
Với : \(a=2^{2018};.b=3^{2019};,c=5^{2020}.\)
Và : \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\Leftrightarrow\)
\(B=1-\frac{1}{2020}< 1< A\)
23 tháng 2 2020
\(M=\frac{2018^{2018}+1}{2019^{2019}+1}\)
\(\Leftrightarrow2M=1+\frac{2017}{2018^{2019}+1}\)
\(N=\frac{2018^{2019}-2}{2018^{2020}-2}\)
\(\Leftrightarrow2N=1-\frac{4034}{2018^{2020}-2}\)
Nhận thấy : \(1+\frac{2017}{2018^{2019}+1}>1-\frac{4034}{2018^{2020}-2}\Leftrightarrow2M>2N\Leftrightarrow M>N\)
W
23 tháng 2 2020
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:)))
22 tháng 5 2019
đặt 22018 = a ; 32019 = b ; 52020 = c
Ta có : \(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
\(B=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(2B=\frac{2}{1.2}+\frac{2}{3.4}+...+\frac{2}{2019.2020}\)
\(< 1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}+\frac{1}{2019.2020}\)
\(2B< 1+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2019-2018}{2018.2019}+\frac{2020-2019}{2019.2020}\)
\(2B< 1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}=1+\frac{1}{2}-\frac{1}{2020}< 1+\frac{1}{2}\)
\(B< \frac{3}{4}\)
\(\Rightarrow A>1>\frac{3}{4}>B\)
22 tháng 5 2019
Mình chỉ biết cách tính B thôi, đây nhé:
B= \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2019.2020}\)
B=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(B=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)
\(B=\frac{1}{1011}+\frac{1}{1012}+....+\frac{1}{2019}+\frac{1}{2020}\)
KN
26 tháng 4 2019
Ta có :
\(N=\frac{2018+2019+2020}{2019+2020+2021}\)
\(=\frac{2018}{2019+2020+2021}+\frac{2019}{2019+2020+2021}+\frac{2020}{2019+2020+2021}\)
Mà \(\frac{2018}{2019}>\frac{2018}{2019+2020+2021}\)
\(\frac{2019}{2020}>\frac{2019}{2019+2020+2021}\)
\(\frac{2020}{2021}>\frac{2020}{2019+2020+2021}\)
\(\Leftrightarrow M>N\)
28 tháng 7 2020
Trả lời:
Ta có:
\(\frac{2018}{2019}>\frac{2018}{2019+2020+2021}\)
\(\frac{2019}{2020}>\frac{2019}{2019+2020+2021}\)
\(\frac{2020}{2021}>\frac{2020}{2019+2020+2021}\)
\(\Rightarrow\frac{2018}{2019}+\frac{2019}{2020}+\frac{2020}{2021}>\frac{2018+2019+2020}{2019+2020+2021}\)
hay \(M>N\)
Vậy \(M>N\)
L
28 tháng 1 2020
\(M=\frac{10^{2018}+1}{10^{2019}+1}\)
\(\Rightarrow10M=\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)
\(N=\frac{10^{2019}+1}{10^{2020}+1}\)
\(\Rightarrow10N=\frac{10\left(10^{2019}+1\right)}{10^{2020}+1}=\frac{10^{2020}+1+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)
Ta co: \(\frac{9}{10^{2019}+1}>\frac{9}{10^{2020}+1}\) ma \(1=1\)
\(\Rightarrow1+\frac{9}{10^{2019}+1}>1+\frac{9}{10^{2020}+1}\)
\(\Rightarrow10M>10N\)
\(\Rightarrow M>N\)
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
1 tháng 10 2019
\(A=\frac{2019^{2020}+1}{2019^{2021}+1}\)và \(B=\frac{2019^{2018}+1}{2019^{2019}+1}\)
Xét \(A=\frac{2019^{2020}+1}{2019^{2021}+1}\Rightarrow2019A=\frac{2019^{2021}+2019}{2019^{2021}+1}=1+\frac{2019}{2019^{2021}+1}\)
Xét \(B=\frac{2019^{2018}+1}{2019^{2019}+1}\Rightarrow2019B=\frac{2019^{2019}+2019}{2019^{2019}+1}=1+\frac{2018}{2019^{2019}+1}\)
Vì \(1+\frac{2018}{2019^{2021}+1}< 1+\frac{2018}{2019^{2019}+1}\Rightarrow\frac{2019^{2020}+1}{2019^{2021}+1}< \frac{2018^{2019}+1}{2019^{2019}+1}\)
\(\Rightarrow A< B\)
9 tháng 2 2020
Ta có:
\(A=\frac{2019^{2020}+1}{2019^{2021}+1}\)
\(\Rightarrow2019A=\frac{2019^{2021}+2019}{2019^{2021}+1}\)
\(\Rightarrow2019A=1+\frac{2019}{2019^{2021}+1}\)
\(\Rightarrow A=1+\frac{2019}{2019^{2021}+1}:2019\)
Ta lại có:
\(B=\frac{2019^{2018}+1}{2019^{2019}+1}\)
\(\Rightarrow2019B=\frac{2019^{2019}+2019}{2019^{2019}+1}\)
\(\Rightarrow2019B=1+\frac{2019}{2019^{2019}+1}\)
\(\Rightarrow B=1+\frac{2019}{2019^{2019}+1}:2019\)
Do \(2019^{2021}+1>2019^{2019}+1\)
\(\Rightarrow\frac{2019}{2019^{2021}+1}< \frac{2019}{2019^{2019}+1}\)
\(\Rightarrow1+\frac{2019}{2019^{2021}+1}:2019< 1+\frac{2019}{2019^{2019}+1}:2019\)
\(\Rightarrow A< B\)
Vậy \(A< B.\)
4 tháng 5 2019
Bạn tham khảo link tại đây nhé :v
https://olm.vn/hoi-dap/detail/217907126396.html
Ta có
\(\frac{2019}{2018}-1=\frac{1}{2018}\)
\(\frac{2020}{2019}-1=\frac{1}{2019}\)
Vì \(\frac{1}{2018}>\frac{1}{2019}\)\(\Rightarrow\)\(\frac{2019}{2018}>\frac{2020}{2019}\)