So sánh C và B 2/ C = 2019^2019 + 1 ——————— 2019^2020 + 1 D = 2019^2020 -2 ——————— 2019^2021 – 2 Mong m.ng

0 bình luận về “So sánh C và B 2/ C = 2019^2019 + 1 ——————— 2019^2020 + 1 D = 2019^2020 -2 ——————— 2019^2021 – 2 Mong m.ng”

1. Ta có: $2019C=\frac{2019^{2020}+2019}{2019^{2020}+1}=\frac{(2019^{2020}+1)+2018}{2019^{2020}+1}=1+\frac{2018}{2019^{2020}+1}$

   $2019D=\frac{2019^{2021}-4038}{2019^{2021}-2}=\frac{(2019^{2021}-2)-4036}{2019^{2021}-2}=1-\frac{4036}{2019^{2021}-2}$

   Vì $1+\frac{2018}{2019^{2020}+1}>1-\frac{4036}{2019^{2021}-2}$

   $⇒2019C>2019D$

   $⇒C>D$

   Vậy $C>D$.

    

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2. Đáp án:

   `C>D`

   Giải thích các bước giải:

   `C=(2019^2019+1)/(2019^2020+1`

   `=>2019C=(2019.(2019^2019+1))/(2019^2020+1)`

   `=>2019C=(2019^2020+2019)/(2019^2020+1`

   `=>2019C=(2019^2020+1+2018)/(2019^2020+1`

   `=>2019C=(2019^2020+1)/(2019^2020+1)+2018/(2019^2020+1`

   `=>2019C=1+2018/(2019^2020+1`

   `D=(2019^2020-2)/(2019^2021-2`

   `=>2019D=(2019.(2019^2020-2))/(2019^2021-2)`

   `=>2019D=(2019^2021-4038)/(2019^2021-2)`

   `=>2019D=(2019^2021-2-4036)/(2019^2021-2`

   `=>2019D=(2019^2021-2)/(2019^2021-2)+(-4036)/(2019^2021-2)`

   `=>2019D=1+(-4036)/(2019^2021-2`

   `=>2018/(2019^2021+1)>(-4036)/(2019^2021-2)`

   `=>1+2018/(2019^2021-1)>1+(-4036)/(2019^2021-2`

   `=>2019C>2019D`

   `=>C>D`

   Vậy `C>D`.

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