So sánh 202010 2020 9 và 202110
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Lời giải:
\(9B=\frac{9^{2019}+9}{9^{2019}+1}=1+\frac{8}{9^{2019}+1}> 1+\frac{8}{9^{2020}+1}=\frac{9^{2020}+9}{9^{2020}+1}=9A\)
$\Rightarrow B>A$
bài 1:
ssh của A là:
(151-3):2+1=75
A=(151+3)x75:2=5775
đáp số: 5775
Ta có: \(2020^{45}-2020^{44}\)
\(=2020^{44}\cdot2020-2020^{44}\cdot1\)
\(=2020^{44}\cdot\left(2020-1\right)\)
\(=2020^{44}\cdot2019\)
Ta có: \(2020^{44}-2020^{43}\)
\(=2020^{43}\cdot2020-2020^{43}\cdot1\)
\(=2020^{43}\cdot\left(2020-1\right)\)
\(=2020^{43}\cdot2019\)
Vì \(2020^{44}>2020^{43}\)
nên \(2020^{44}\cdot2019>2020^{43}\cdot2019\)
hay \(2020^{45}-2020^{44}>2020^{44}-2020^{43}\)
`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`
Giải:
Ta có: N=2019+2020/2020+2021
=>N=2019/2020+2021 + 2020/2020+2021
Vì 2019/2020 > 2019/2020+2021 ; 2020/2021 > 2020/2020+2021
=>M>N
Vậy ...
Chúc bạn học tốt!
Ta có : \(\dfrac{2019}{2020}>\dfrac{2019}{2020+2021}\)
\(\dfrac{2020}{2021}>\dfrac{2020}{2020+2021}\)
\(\Rightarrow\dfrac{2019}{2020}+\dfrac{2020}{2021}>\dfrac{2019+2020}{2020+2021}\)
\(\Rightarrow M>N\)
20201020209>202110