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Ta có:
\(A=\frac{2018+2019}{2019+2020}=\frac{2018}{2019+2020}+\frac{2019}{2019+2020}\)
mà 2019+2020 >2019>2020 \(\Rightarrow\frac{2018}{2019+2020}< \frac{2018}{2019};\frac{2019}{2019+2020}< \frac{2019}{2020}\)
\(\Rightarrow\frac{2018}{2019+2020}+\frac{2019}{2019+2020}< \frac{2018}{2019}+\frac{2019}{2020}\)hay \(A< B\)
Ta có:\(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times..\times\frac{2018}{2019}\times\frac{2019}{2020}\)\(=\frac{1}{2020}\)
Vậy biểu thức \(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times..\times\frac{2018}{2019}\times\frac{2019}{2020}\)\(=\frac{1}{2020}\)
1/2 x 2/3 x 3/4 x ... x 2018/2019 x 2019/2020
= 1 x 2 x 3 x ... x 2018 x 2019 / 2 x 3 x 4 x ... x 2019 x 2020
Khử loại đi ta còn lại phân số 1/2020
Hok tốt ^^
tớ nghĩ là các phân số trên đều là những ps nhỏ hơn 1 lên A<3
mình chỉ nghĩ thôi. K biết đúng hay sai đâu. đúng thì tích còn sai thì bỏ qua cho
Trả lời :...............................................
\(\frac{4078379}{4078379}\)
Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
k nhé Kim Râu La
\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018\times2019}\)
\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(A=\left(\dfrac{2020}{2019}-\dfrac{1}{2019}\right)-\left(\dfrac{2019}{2018}-\dfrac{1}{2018}\right)\)
\(A=\left(\dfrac{2020-1}{2019}\right)-\left(\dfrac{2019-1}{2018}\right)\)
\(A=1-1\)
\(A=0.\)
\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018\times2019}\)
\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(A=\left(\dfrac{2020}{2019}-\dfrac{1}{2019}\right)-\left(\dfrac{2019}{2018}-\dfrac{1}{2018}\right)\)
\(A=\dfrac{2019}{2019}-\dfrac{2018}{2018}\)
\(A=1-1\)
\(A=0\)
https://olm.vn/hoi-dap/detail/224964577156.html
THAM-KHẢO-NHÉ
THANKS
Ta có: \(\frac{2018}{2019}\)+ \(\frac{2019}{2020}\)+\(\frac{2020}{2018}\)= (1-\(\frac{1}{2019}\)) + ( 1 -\(\frac{1}{2020}\)) + ( 1 - \(\frac{1}{2018}\)) = ( 1+1+1) - (\(\frac{1}{2019}+\frac{1}{2020}+\frac{1}{2018}\)) = 3 - (\(\frac{1}{2019}+\frac{1}{2020}+\frac{1}{2018}\)) \(\Leftrightarrow\)3 - (\(\frac{1}{2019}+\frac{1}{2020}+\frac{1}{2018}\)) <3 Vậy \(\frac{2018}{2019}+\frac{2019}{2020}+\frac{2020}{2018}\)< 3
\(A=2018\times2020+2021\) và \(B=2019\times2019+2021\)
\(A=2018\times2019+2018+2021\)
\(B=2018\times2019+2019+2021\)
Vì \(2019>2018\Rightarrow A< B\)
Ta có:
\(1-\frac{2017}{2018}=\frac{1}{2018};1-\frac{2018}{2019}=\frac{1}{2019};1-\frac{2019}{2020}=\frac{1}{2020}\)
Vì \(\frac{1}{2018}>\frac{1}{2019}>\frac{1}{2020}\)nên \(\frac{2017}{2018}< \frac{2018}{2019}< \frac{2019}{2020}\)
2017/2018 = (2018-1)/2018 = 1-1/2018
2018/2019 = (2019-1)/2019 = 1 - 1/2019
2019/2020 = (2020-1)/2020 = 1 - 1/2020
Có 1/2018 > 1/2019 > 1/2020 => 2017/2018 < 2018/2019 < 2019/2020
NMD yêu mn
Đề hỏi gì ??
_Minh ngụy_