Tính nhanh: A = $\frac{2020^{3}+1}{2020^{2}-2019}$ B = $\frac{2020^{3}-1}{2020^{2}+2021}$

Question

Tính nhanh: A = $\frac{2020^{3}+1}{2020^{2}-2019}$
B = $\frac{2020^{3}-1}{2020^{2}+2021}$

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Ladonna 4 years 2020-10-30T01:36:00+00:00 2 Answers 76 views 0

Answers ( )

  1. Philomena
    0
    2020-10-30T01:37:29+00:00
    Reply

     A=2020^3+1/2020^2-2019

    A=(2020+1)(2020^2-2020+1)/2020^2-2019

    A=2021.(2020^2-2019)/2020^2-2019

    A=2021.

    B=2020^3-1/2020^2+2021

    B=(2020-1)(2020^2+2020+1)/2020^2+2021

    B=2019.(2020^2+2021)/2020^2+2021

    B=2019

  2. Acacia
    0
    2020-10-30T01:37:55+00:00
    Reply

    Đáp án:

    $A=2021$

    $B=2019$

    Giải thích các bước giải:

    $A=\dfrac{2020^3+1}{2020^2-2019}$

    $=\dfrac{(2020+1)(2020^2-2020+1)}{2020^2-2019}$

    $=\dfrac{2021.(2020^2-2019)}{2020^2-2019}$

    $=2021$

    $B=\dfrac{2020^3-1}{2020^2+2021}$

    $=\dfrac{(2020-1)(2020^2+2020+1}{2020^2+2021}$

    $=\dfrac{2019.(2020^2+2021)}{2020^2+2021}$

    $=2019$

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