Ta có 2021² = 2021 . 2021

Vì 2020 < 2021 nên 2021 . 2020 < 2021 . 2021 hay 2021 . 2020 < 2021²

\(\dfrac{5^{2021}}{5^{2020}}\cdot5^2=5\cdot5^2=5^3\)

5²⁰²¹:5²⁰²⁰. 5²

= 5¹.5²

=5³

Đặt \(2020-x=u;x-2021=v\)thì \(u+v=-1\)

Phương trình trở thành \(\frac{u^2+uv+v^2}{u^2-uv+v^2}=\frac{19}{49}\Leftrightarrow30u^2+30v^2+68uv=0\)

\(\Leftrightarrow15\left(u+v\right)^2+4uv=0\Leftrightarrow4uv=-15\Leftrightarrow uv=\frac{-15}{4}\)

hay \(\left(2020-x\right)\left(x-2021\right)=-\frac{15}{4}\Leftrightarrow x^2-4041x+4082416,25=0\)

Dùng công thức nghiệm tìm được x = 2022, 5 hoặc x = 2018, 5

Ta có A = \(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{2021}\)

= \(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2021}}\)

=> 2A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}\)

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

=> A = \(\frac{1}{2}-\frac{1}{2^{2021}}< \frac{1}{2}\left(\text{ĐPCM}\right)\)

Sửa đề: \(S=1+3+3^2+3^3+\cdots+3^{2021}+3^{2022}\)

Ta có: \(S=1+3+3^2+3^3+\cdots+3^{2021}+3^{2022}\)

\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5+3^6\right)+\left(3^7+3^8+3^9+3^{10}\right)+\cdots+\left(3^{2019}+3^{2020}+3^{2021}+3^{2022}\right)\)

\(=13+3^3\left(1+3+3^2+3^3\right)+3^7\left(1+3+3^2+3^3\right)+...+3^{2019}\left(1+3+3^2+3^3\right)\)

\(=13+40\left(3^3+3^7+\cdots+3^{2019}\right)=3+10+10\cdot4\cdot\left(3^3+3^7+\cdots+3^{2019}\right)\)

=>S chia 10 dư 3

=>S có tận cùng là 3

\(\left(5^{2021}-5^{2020}\right):5^{2020}=5^{2021}:5^{2020}-5^{2020}:5^{2020}=5-1=4\)

làm ơn đi, mình đang cần gắp.