So sánh : A= $\frac{2020^{2018}-1 }{2020^{2019}-2019 }$ và B=$\frac{2020^{2019}+1 }{2020^{2020}+2019 }$

0 bình luận về “So sánh : A= $\frac{2020^{2018}-1 }{2020^{2019}-2019 }$ và B=$\frac{2020^{2019}+1 }{2020^{2020}+2019 }$”

1. Tham khảo

    Xét `A`

   `⇒2020A=\frac{2020^{2019}-2020}{2020^{2019}-2019}`

   `⇒2020A=1-\frac{1}{2020^{2019}-2019}`

   `⇒2020A<1(1)`

   Xét `B`

   `⇒2020B=\frac{2020^{2020}+2020}{2020^{2020}+2019}`

   `⇒2020B=1+\frac{1}{2020^{2020}+2019}`

   `⇒2020B>1(2)`

   Từ `(1)(2)⇒2020A<2020B`

   `⇒A<B`

   `\text{©CBT}`

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2. So sánh: `A={2020^2018-1}/{2020^2019-2019}`

         và `B={2020^2019+1}/{2020^2020+2019}`
   `+)`Ta có: `A={2020^2018-1}/{2020^2019-2019}`
    `=> 2020A={2020^2019-2020}/{2020^2019-2019}`
   `<=> 2020A={2020^2019-2019-1}/{2020^2019-2019}`
   `<=> 2020A=1+{-1}/{2020^2019-2019}`
   Vì `{-1}/{2020^2019-2019} <0` nên `1+{-1}/{2020^2019-2019} <1`
   `=> A < 1`
   `+)`Ta có `B={2020^2019+1}/{2020^2020+2019}`
    `=>2020B={2020^2020+2020}/{2020^2020+2019}`
   `<=>2020B={2020^2020+2019+1}/{2020^2020+2019}`
   `<=>2020B=1+1/{2020^2020+2019}`
   Vì `1/{2020^2020+2019} > 0` nên `1+1/{2020^2020+2019} > 1`
   Vì `2020A<1 ; 2020B>1`
   `=> A<B`

   #Chucbanhoctot

   #XinCTLHN

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