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#)Giải :
\(Q=2+\frac{2016}{2017+2018+2019}+\frac{2017}{2017+2018+2019}+\frac{2018}{2017+2018+2019}\)
Ta thấy : \(2>\frac{2016}{2017};2>\frac{2017}{2018};2>\frac{2018}{2019}\left(1\right)\)
\(\frac{2016}{2017+2018+2019}< \frac{2016}{2017}\left(2\right)\)
\(\frac{2017}{2017+2018+2019}< \frac{2017}{2018}\left(3\right)\)
\(\frac{2018}{2017+2018+2019}< \frac{2018}{2019}\left(4\right)\)
Từ (1) (2) (3) (4) \(\Rightarrow P>Q\)
\(\frac{2016}{2017}\)x \(\frac{2017}{2018}\)x \(\frac{2019}{2020}\)=\(\frac{504}{505}\)
đ/s:\(\frac{504}{505}\)
\(\frac{2016}{2017}\times\frac{2017}{2018}\times\frac{2018}{2019}\times\frac{2019}{2020}\)=
\(0,998109801980198\)
Đổi ra ta sẽ có !
\(\frac{504}{505}\)
Vậy là : ...................
\(A=\frac{2017.2018-1}{2017.2018}=1-\frac{1}{2017.2018}\)(1)
\(B=\frac{2018.2019-1}{2018.2019}=1-\frac{1}{2018.2019}\)(2)
Từ(1) và (2)
\(\Rightarrow B>A\)
Ta có:
2017*2018+2018*2019=2018*(2017+2019)=2018*4036=8144648
2018*9-2017*9=9*(2018-2017)=9*1=9
Vậy 8144648*9=73301832
\(A=\frac{2017\times2108}{2017\times2018+1}=1-\frac{1}{2017\times2018+1}\)
\(B=\frac{2018\times2019}{2018\times2019+1}=1-\frac{1}{2018\times2019+1}\)
Nhận thấy:\(2017\times2018< 2018\times2019\)
=> \(2017\times2018+1< 2018\times2019+1\)
=> \(\frac{1}{2017\times2018+1}>\frac{1}{2018\times2019+1}\)
=> \(A< B\)
`a,`
`5/6=1-1/6`
`7/8=1-1/8`
Mà `1/6>1/8 -> 5/6<7/8`
`b,`
`9/5=(9 \times 2)/(5 \times 2)=18/10`
`3/2=(3 \times 5)/(2 \times 5)=15/10`
`18/10 > 15/10 -> 9/5 > 3/2`
`c,`
`2017/2018 = 1-1/2018`
`2019/2020=1-1/2020`
`1/2018 > 1/2020 -> 2017/2018 < 2019/2020`
`d,`
`2018/2017 = 1+1/2017`
`2020/2019 = 1+1/2019`
`1/2017 > 1/2019 -> 2018/2017>2020/2019`
A= 2017/2018 + 2018/2019 + 2019/2017
= 3.0000007368
Nếu sai mog các bạn đừng gạch đá !