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Lời giải:
$3^{2022}+3^{2020}-(2^{2020}+2^{2020})$

$=3^{2020}(3^2+1)-2.2^{2020}=10.3^{2020}-2^{2021}$

Ta thấy: $10.3^{2020}\vdots 10$, còn $2^{2021}\not\vdots 10$ nên $10.3^{2020}-2^{2021}\not\vdots 10$ 

Bạn xem lại đề.

2A=2*(1+2+2²+...+2²⁰²⁰)=2+2²+...+2²⁰²¹

^2A-A=(1+2+2²+...+2²⁰²¹)-(1+2+2²+...+2²⁰²⁰)

A=2²⁰²¹-1<2021

Giải:

A=1+2+2²+2³+...+2²⁰²⁰

2A=2+2²+2³+2⁴+...+2²⁰²¹

2A-A=(2+2²+2³+2⁴+...+2²⁰²¹)-(1+2+2²+2³+...+2²⁰²⁰)

A=2²⁰²¹-1

⇒A<2²⁰²¹

Chúc bạn học tốt!

Ta có 106 - 57 = 26 . 56 - 57

= 56 . (26 - 5)

= 56 . (64 - 5)

= 56 . 59 chia hết cho 59

Vậy 106 - 57 chia hết cho 59.

tik nha!

ta có\(\Delta=\left(4^2\right)-4\cdot-1\cdot-6=-120\)

=> vô nghiệm

\(=2+1=3\)

2²⁰²¹:2²⁰²⁰+2²⁰²²:2²⁰²⁰

         1         +        2

                   3

đây chứ đâu

\(\left(x-4\right)\cdot2^{2020}=2^{2022}\)

\(\Rightarrow\left(x-4\right)\cdot2^{2020}-2^{2022}=0\)

\(\Rightarrow2^{2020}\cdot\left[\left(x-4\right)-2^2\right]=0\)

\(\Rightarrow2^{2020}\cdot\left(x-4-4\right)=0\)

\(\Rightarrow2^{2020}\cdot\left(x-8\right)=0\)

\(\Rightarrow x-8=0\)

\(\Rightarrow x=8\)

\(A=\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{199}+\dfrac{1}{200}\\ >\dfrac{1}{10}+\left(\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}\right)\left(90so\right)+\left(\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}\right)\left(100so\right)\\ A>\dfrac{1}{10}+\dfrac{90}{100}+\dfrac{100}{200}=1+\dfrac{1}{2}=\dfrac{3}{2}\left(đpcm\right).\)

\(M=2^{2020}-2^{2020}+1=1\)