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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có 20212 = 2021 . 2021
Vì 2020 < 2021 nên 2021 . 2020 < 2021 . 2021 hay 2021 . 2020 < 20212
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 407 + 24 + (-407) + (-84)
= [407 + (-407)] + 24 + (-84)
= 0 + 24 + (-84)
= 24 + (-84)
= (-60)
b) (-42) x 23 + 77 x (-42) - 200
= (-42) x (23 + 77) - 200
= (-42) x 100 - 200
= (-4200) - 200
= (-4400)
c) (-2021) + (499 + 2021) + 41
= (-2021) + 499 + 2021 + 41
= [(-2021) + 2021] + 499 + 41
= 0 + 499 + 41
= 499 + 41
= 540
d) (-202) - (34 - 202) - 66 + 50
= (-202) - 34 + 202 - 66 + 50
= [(-202) + 202] - 34 - 66 + 50
= 0 - 34 - 66 + 50
= (-34) - 66 + 50
= (-100) + 50
= (-50)
e) Bạn viết lại đề bài.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\left|x-2021\right|\ge0\forall x\)
\(\Leftrightarrow2\left|x-2021\right|\ge0\forall x\)
\(\Leftrightarrow2\left|x-2021\right|+9\ge9\forall x\)
Dấu '=' xảy ra khi x=2021
b) Ta có: \(\left|x-2\right|\ge0\forall x\)
\(\left|y+1\right|\ge0\forall y\)
Do đó: \(\left|x-2\right|+\left|y+1\right|\ge0\forall x,y\)
\(\Leftrightarrow\left|x-2\right|+\left|y+1\right|+2021\ge2021\forall x,y\)
Dấu '=' xảy ra khi (x,y)=(2;-1)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
B/A
\(=\dfrac{1+\dfrac{2020}{2}+1+\dfrac{2019}{3}+...+1+\dfrac{1}{2021}+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}\)
\(=\dfrac{2022\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}=2022\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(2021-1\right)\left(2021-2\right)\cdot\left(2021-3\right)\cdot...\cdot\left(2021-n\right)\)
Tích trên có đúng 2021 thừa số nên n=2021
=>\(A=\left(2021-1\right)\left(2021-2\right)\cdot\left(2021-3\right)\cdot...\cdot\left(2021-2021\right)\)
\(=2020\cdot2019\cdot2018\cdot...\cdot0\)
=0