Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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=1-1/2019+1-1/2020+1+2/2018
=>A=(1+1+1)+(1/2018-1/2009)+(1/2018-1/2020)
Vì 1/2018>1/2019 và 1/2028>1/2020
=>A>3
Vậy a >A
study well
k nha ủng hộ mk nhé
Mình cũng làm giống thế . nhưng con bạn mình làm a < 3 nên mình không chắc chắn
\(\frac{2017}{2018}\)và \(\frac{2019}{2020}\)
Ta có : \(1-\frac{2017}{2018}=\frac{1}{2018};1-\frac{2019}{2020}=\frac{1}{2020}\)
Vì \(\frac{1}{2018}>\frac{1}{2020}\)nên \(\frac{2017}{2018}< \frac{2019}{2020}\)
Cái này là so sánh bằng phần bù của đơn vị nha bn !
Học tốt #
\(\frac{2017}{2018};\frac{2018}{2019};\frac{2019}{2020}\)
\(\Rightarrow\frac{2017}{2018}< \frac{2019}{2020}\)
a: Số cần tìm là 5,32:0,125=42,56
b: \(A=1+\dfrac{1}{2019}-1-\dfrac{1}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}=0\)
vi 2018/2019<1
2019/2020<1
2020/2021<1
nen 2018/2019 + 2019/2020 + 2020/2021<1+1+1=3
Ta có : \(\frac{1}{n}+\frac{2020}{2019}=\frac{2019}{2018}+\frac{1}{n+1}\)
=> \(\frac{1}{n}-\frac{1}{n+1}=\frac{2019}{2018}-\frac{2020}{2019}\)
=> \(\frac{n+1}{n\left(n+1\right)}-\frac{n}{\left(n+1\right)n}=\frac{1}{4074342}\)
=> \(\frac{1}{n\left(n+1\right)}=\frac{1}{2018.2019}\)
=> n(n + 1) = 2018.2019
=> n(n + 1) = 2018.(2018 + 1)
=> n = 2018
Ok em, để olm.vn giúp em nhá:
A = \(\dfrac{1}{2}\):3 + \(\dfrac{1}{3}\):4 + \(\dfrac{1}{4}\):5+...+\(\dfrac{1}{2018}\):2019 + \(\dfrac{1}{2019}\): 2020
A=\(\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+\dfrac{1}{4}\times\dfrac{1}{5}+..+\dfrac{1}{2018}\times\dfrac{1}{2019}+\dfrac{1}{2019}\times\dfrac{1}{2020}\)
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+....+ \(\dfrac{1}{2018}\) - \(\dfrac{1}{2019}\)+ \(\dfrac{1}{2019}\) - \(\dfrac{1}{2020}\)
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{2020}\)
A = \(\dfrac{1009}{2020}\)
\(\left(2020\frac{2018}{2021}-2019\frac{20182018}{20212021}\right):\frac{2018}{2021}\)
\(=\left(2020\frac{2018}{2021}-2019\frac{2018}{2021}\right):\frac{2018}{2021}\)
\(=1:\frac{2018}{2021}=\frac{2021}{2018}\)
\(A=\frac{2020}{2019}-\frac{2019}{2018}+\frac{1}{2019\times2018}\)
\(=\frac{2020\times2018}{2019\times2018}-\frac{2019\times2019}{2019\times2018}+\frac{1}{2019\times2018}\)
\(=\frac{2020\times2018-2019\times2019+1}{2019\times2018}\)
\(=\frac{\left(2019+1\right)\times\left(2019-1\right)-2019\times2019+1}{2019\times2018}\)
\(=\frac{2019\times2019-2019+2019-1-2019\times2019+1}{2019\times2018}\)
\(=\frac{2019\times2019-1-\left(2019\times2019-1\right)}{2019\times2018}\)
\(=\frac{0}{2019\times2018}\)
\(=0\)
Vậy A = 0
ta có
A=2020*2018/2019*2018-2019*2019/2018*2019+1/2018*2019
=>A*(2018*2019)=2020*2018-2019*2019+1
=>A*(2018*2019)=(2019+1)*2018-(2018+1)*2019+1
=>A*(2018*2019)=(2019*2018+2018)-(2018*2019+2019)+1
=>A*(2018*2019)=2019*2018+2018-2018*2019-2019+1
=>A*(2018*2019)=2018-2019+1
=>A*(2018*2019)=2018+1-2019
=>A*(2018*2019)=0
=>A=0/(2018*2019)
=>A=0