help me !!!!!!!

\(=2021\cdot2\cdot\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)=4042\cdot\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)=0\)

𝑝=−2856279824648840

= 1.495676488x10^-3

(1+3+5+7+...+2019+2021)

A=1−3+5−7+......−2019+2021−2023

A=(1−3)+(5−7)+....+(2021−2023)A=(1−3)+(5−7)+....+(2021−2023)

A=−2+(−2)+....+(−2)(506)A=−2+(−2)+....+(−2)(506cặp)

a=−2.506A=−2.506

A=−1012A=−1012

(2+4+6+8+...+2020)

B=2+4+6+8+...+2018+2020
B = 2(1 + 2 + 3 + 4 + ... + 1009 + 1010)
B = 2 . (1011 . 1010 : 2)
B = 2 . 510555
B = 1 021 110

(1+3+5+7+......+2019+2021)-(2+4+6+8+.....+2020)

\(=\dfrac{\left(1+2021\right).\left[\left(2021-1\right):2+1\right]}{2}-\dfrac{\left(2+2020\right).\left[\left(2020-2\right):2+1\right]}{2}\)

\(=1011\)

A = B

a)

\(A=\frac{2020^3+1}{2020-2019}=\frac{\left(2020+1\right)\left(2020^2-2020+1\right)}{2020-2020+1}\) \(=2020+1=2021\)

b)

B = \(\frac{2020^3-1}{2020^2+2021}=\frac{\left(2020-1\right)\left(2020^2+2020+1\right)}{2020^2+2020+1}\) \(=2020-1=2019\)

a. \(A=\frac{2020^3+1}{2020^2-2019}=\frac{\left(2020+1\right)\left(2020^2-2020+1\right)}{2020^2-2020+1}=2020+1=2021\)

b. \(B=\frac{2020^3-1}{2020^2+2021}=\frac{\left(2020-1\right)\left(2020^2+2020+1\right)}{2020^2+2020+1}=2020-1=2019\)

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Ta có: \(2021^2=\left(2020+1\right)^2=2020^2+2.2020.1+1^2\)

\(\Rightarrow1+2020^2=2021^2-2.2020\)

\(\Rightarrow\sqrt{1+2020^2+\frac{2020^2}{2021}}+\frac{2020}{2021}\)

\(=\sqrt{2021^2-2.2020+\frac{2020^2}{2021}}+\frac{2020}{2021}\)

\(=\sqrt{2021^2-2.2021.\frac{2020}{2021}+\left(\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)

\(=\sqrt{\left(2021-\frac{2020}{2021}\right)^2}+\frac{2020}{2021}\)

\(=2021-\frac{2020}{2021}+\frac{2020}{2021}=2021\)