\(S=1+2+2^2+2^3+2^4+...+2^{2011}\)

\(\Rightarrow S=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{2009}\left(1+2+2^2\right)\)

\(\Rightarrow S=7+2^3.7+...+2^{2009}.7\)

\(\Rightarrow S=7\left(1+2^3+...+2^{2009}\right)⋮7\)

\(\Rightarrow dpcm\)

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!

\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\)

\(\Rightarrow\dfrac{1}{2}A=\dfrac{1}{2}.\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)\)\(\Rightarrow\dfrac{1}{2}A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\)

\(\Rightarrow A-\dfrac{1}{2}A=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\right)\)\(\Rightarrow\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^{2022}}\)

\(\Rightarrow\dfrac{1}{2}A=\dfrac{2^{2021}-1}{2^{2022}}\)

\(\Rightarrow A=\dfrac{2^{2021}-1}{2^{2023}}.2=\dfrac{2^{2021}-1}{2^{2021}}\)

Vậy \(A=\dfrac{2^{2021}-1}{2^{2021}}\)

A=1/2+1/22+1/23+...+1/22020+1/22021 > B=1/3+1/4+1/5+13/60

Ta có: $A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2021}}+\frac{1}{2^{2022}}$

$\Rightarrow 2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}+\frac{1}{2^{2021}}$

$\Rightarrow 2A-A=(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2020}}+\frac{1}{2^{2021}})-(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{2021}}+\frac{1}{2^{2022}})$

$\Rightarrow A=1-\frac{1}{2^{2022}}<1$

$\Rightarrow A<1\left(1\right)$

Lại có: $B=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{17}{60}$

$\Rightarrow B=\frac{16}{15}$

$\Rightarrow B=1+\frac{1}{15}>1$

$\Rightarrow B>1\left(2\right)$

Từ $\left(1\right)$ và $\left(2\right)\Rightarrow A<B$

Vậy $A<B$

kp[

\(A=2+2^2+2^3+...+2^{2020}+2^{2021}+2^{2022}\\=(2+2^2)+(2^3+2^4)+(2^5+2^6)+...+(2^{2021}+2^{2022})\\=2\cdot(1+2)+2^3\cdot(1+2)+2^5\cdot(1+2)+...+2^{2021}\cdot(1+2)\\=2\cdot3+2^3\cdot3+2^5\cdot3+...+2^{2021}\cdot3\\=3\cdot(2+2^3+2^5+..+2^{2021})\)

Vì \(3\cdot\left(2+2^3+2^5+...+2^{2021}\right)⋮3\)

nên \(A⋮3\).

\(Toru\)

A=(2+2²)+2²(2+2²)+...+2²⁰²⁰(2+2²)

A= 6.1+2².6+...+2²⁰²⁰.6

A=6(1+2²+...+2²⁰²⁰) chia hết cho 3

vậy A chia hết cho 3

\(A=1+2^1+2^2+2^3+...+2^{2021}\\2A=2+2^2+2^3+2^4+...+2^{2022}\\2A-A=(2+2^2+2^3+2^4+...+2^{2022})-(1+2^1+2^2+2^3+...+2^{2021})\\A=2^{2022}-1\\\Rightarrow A+1=2^{2022}\)

Mặt khác: \(2^x=A+1\)

\(\Rightarrow 2^x=2^{2022}\\\Rightarrow x=2022(tm)\)

Vậy x = 2022.

Bài 2: 

a: =>x-45=-20

hay x=25

b: =>35+30-5x=-12+112

=>65-5x=100

=>5x=-35

hay x=-7

c: =>x-124=1000

hay x=1124

d: \(\Leftrightarrow46-\left(3x-2\right)^3=-18\)

\(\Leftrightarrow\left(3x-2\right)^3=64\)

=>3x-2=4

hay x=2