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#)Giải :
Ta có : \(A=\frac{2017}{2018}+\frac{2018}{2019}+\frac{2019}{2019}< 1+1+1\)
\(\Rightarrow A< 3\)
Mình giải thế này cho ngắn gọn, với lại nhanh ^^
Phân tích 2 phân số ta có:
1 = \(\dfrac{2017\times2019}{2017\times2019}\) = \(\dfrac{\left(2018-1\right)\times\left(2018+1\right)}{2017\times2019}\) = \(\dfrac{2018^2-1^2}{2017\times2019}\)
\(\dfrac{2018\times2018}{2017\times2019}\) = \(\dfrac{2018^2}{2017\times2019}\)
Vì \(2018^2\) > \(2018^2-1^2\) nên \(\dfrac{2018^2}{2017\times2019}\) > \(\dfrac{2018^2-1^2}{2017\times2019}\) hay \(\dfrac{2018\times2018}{2017\times2019}\) > 1
(Áp dụng hằng đẳng thức \(a^2-b^2\) = (a - b)(a + b))
nhầm dòng 2 nhé
\(=\dfrac{2018\times2018}{2018\times2018-1}=\)
Vì \(2018\times2018>2018\times2018-1\) nên \(\dfrac{2018\times2018}{2018\times2018-1}>1\)
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 \(A< B\)
#)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) \(\Rightarrow P>Q\)
`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 nhỏ hơn B
Ta có :
\(\frac{2018}{2019}\)\(+\)\(\frac{2019}{2018}\)\(=\frac{2018}{2019}\)\(+\frac{1}{2018}\)\(+1>\frac{2018+1}{2019}\)\(+1\)
\(=1+1=2\)
\(\Rightarrow\frac{2018}{2019}\)\(+\frac{2019}{2018}\)\(>2\)
\(\Rightarrow A>B\)