a)Có A=5+5²+5³+...+5²⁰¹⁶

=>5A=5²+5³+...+5²⁰¹⁷

=>4A=5A-A=5²⁰¹⁷-5

=>4A+5=5²⁰¹⁷-5+5=5²⁰¹⁷=5^x

=>x=2017

b) Gọi 4 số tự nhiên liên tiếp là : k;k+1;k+2;k+3

Có k(k+1)(k+2)(k+3)+1

=k(k+3)(k+1)(k+2)+1

=(k²+3k)(k²+3k+2)+1

Đặt k²+3k=A

=A(A+2)+1

=A²+2A+1

=(A+1)²

=>ĐPCM

Tìm GTNN của biểu thức sau: A=4x^2-4xy+5y^2+20x-6y+2044

\(A=5+5^2+5^3+...+5^{2016}\)

\(5A=5^2+5^3+5^4+...+5^{2017}\)

\(\rightarrow5A-A=5^{2017}-5\)

\(4A=5^{2017}-5\)

\(\Rightarrow4A+5=5^{2017}-5+5\)

Mà \(4A+5=5^x\)

\(\Rightarrow5^x=5^{2017}\)

Vậy \(x=2017\)

a/ \(A=5+5^2+5^3+..........+3^{2016}\)

\(\Leftrightarrow A=\left(5+5^4\right)+\left(5^2+5^5\right)+...........+\left(5^{2013}+5^{2016}\right)\)

\(\Leftrightarrow A=5\left(1+5^3\right)+5^2\left(1+5^3\right)+..........+5^{2013}\left(1+5^3\right)\)

\(\Leftrightarrow A=5.126+5^2.126+............+5^{2013}.126\)

\(\Leftrightarrow A=126\left(1+5^2+........+5^{2013}\right)⋮126\left(đpcm\right)\)

b/ \(A=5+5^2+5^3+..........+5^{2016}\)

\(\Leftrightarrow5A=5^2+5^3+...............+5^{2016}+5^{2017}\)

\(\Leftrightarrow5A-A=\left(5^2+5^3+........+5^{2017}\right)-\left(5+5^2+.......+5^{2016}\right)\)

\(\Leftrightarrow4A=5^{2017}-5\)

\(\Leftrightarrow4A+5=5^{2017}\)

\(\Leftrightarrow4A+5\) là 1 lũy thừa

c/ Ta có :

\(4A+5=5^{2017}\)

Mà \(4A+5=5^x\)

\(\Leftrightarrow5^{2017}=5^x\)

\(\Leftrightarrow x=2017\)

Vậy ..

\(A=5+5^2+5^3+...+5^{2022}\)

\(5A=5^2+5^3+5^4+...+5^{2023}\)

\(5A-A=5^{2023}-5\)

\(4A+5=5^{2023}-5+5\)

\(4A+5=5^{2023}\)

Vì \(4A+5=5^x\)

\(=>5^x=5^{2023}\)

\(=>x=2023\)

\(#PaooNqoccc\)

A = 5 + 5² + 5³ + ... + 5²⁰²⁰

⇒ 5A = 5² + 5³ + 5⁴ + ... + 5²⁰²¹

⇒ 4A = 5A - A

= (5² + 5³ + 5⁴ + ... + 5²⁰²¹) - (5 + 5² + 5³ + ... + 5²⁰²⁰)

= 5²⁰²¹ - 5

⇒ 4A + 5 = 5²⁰²¹ - 5 + 5

= 5²⁰²¹

Mà 4A + 5 = 5ˣ

5ˣ = 5²⁰²¹

x = 2021

A=5+5^2+5^3+...+ 5^2020
5A= 5^2 +5^3+...+5^2021
5A-A= _ 5^2+5^3+...+5^2021
              5+5^2+5^3+...+5^2020
             ____________________
   4A=     5^2021 - 5
Vậy 4A+5=5^x
5^2021-5+5=5^x
5^2021-5    = 5^x - 5
Chiệt tiêu - 5 ở hai bên đi ta còn:
5^2021=5^x
=> x=2021
 

hay

A=5+5^2+...+5^2017
5A=5^2+5^3+5^4+.....+5^2018
4A=5^2+5^3+..+5^2018-5-5^2-5^3-...-5^2017
4A=5^2018-5
=>4A+5=5^x
5^2018-5+5=5^x
5^2018=5^x
=>x=2018

nhớ k nha

Ta có:

A=5+5²+5³+...+5²⁰¹⁷

5A=5²+5³+...+5²⁰¹⁷+5²⁰¹⁸

4A=5²⁰¹⁸-5

4A+5=5²⁰¹⁸

5^x=5²⁰¹⁸

x=2018

Ta có:A=5+5²+5³+...+5²⁰¹⁷

5A=5²+5³+5⁴+...+5²⁰¹⁸

\(\Rightarrow\)5A-A=(5²+5³+5⁴+...+5²⁰¹⁸)-(5+5²+5³+...+5²⁰¹⁷)

\(\Rightarrow\)4A=5²⁰¹⁸-5

\(\Rightarrow\)4A+5=5²⁰¹⁸

\(\Rightarrow\)5^x=5²⁰¹⁸

\(\Rightarrow\)x=2018

Vậy x=2018.

câu hỏi tương tự

A=5+5²+...+5²⁰¹⁶

5A=5²+5³+...+5²⁰¹⁷

5A-A=(5²+5³+...+5²⁰¹⁷)-(5+5²+...+5²⁰¹⁶)

4A = 5²⁰¹⁷ - 5

=> 4A + 5 = 5²⁰¹⁷ - 5 + 5 = 5²⁰¹⁷ = 5^n-1

=> n-1=2017 => n=2018

A = 5 + 5² + 5³ + ............ + 5²⁰¹⁶

5A = 5² + 5³ + 5⁴ + .............. + 5²⁰¹⁷

5A - A = ( 5² + 5³ + 5⁴ + ................ + 5²⁰¹⁷ ) - ( 5 + 5² + 5³ + ................. + 5²⁰¹⁶ )

5A - A = 5² + 5³ + 5⁴ + ........... + 5²⁰¹⁷ - 5 - 5² - 5³ - .............. - 5²⁰¹⁶

4A = 5²⁰¹⁷ - 5

4A + 5 = 5^n-1

\(\Rightarrow\) 4A + 5 = 5²⁰¹⁷ - 5 + 5 = 5²⁰¹⁷ = 5^n-1

\(\Rightarrow\) n - 1 = 2017 

\(\Rightarrow\) n = 2018

Vậy n = 2018

A=5.5^2+...+5^2016

5A=5^2+5^3+...+2017

5A-A=(5^2+5^3+...+5^2017)-(5+5^2+5^3+...+5^2016)

4A=5^2017-5

=>4A+5=5^2017-5+5 =5^2017=5^n-1

=>n-1=2017=>n=2018

mk ko biết đúng hay sai nhé

Ta có :\(5A=5^2+5^3+5^4+...+5^{2018}\)

\(\Rightarrow5A-A=\left(5^2+5^3+5^4+...+5^{2018}\right)-\left(5+5^2+5^3+...+5^{2017}\right)\)

\(\Rightarrow4A=5^{2018}-5\)

Theo đề bài : \(25^x=4A+5\Leftrightarrow25^x=5^{2018}\)

\(\Leftrightarrow5^{2x}=5^{2018}\Leftrightarrow2x=2018\Leftrightarrow x=2014\)

ta tính A=5+5^2+5^3+....+5^2016+5^2017

5A=5^2+5^3+...+5^2017+5^2018

5A-A=(5^2+5^3+...+5^2017+5^2018)-(5+5^2+...+5^2017)

4A=5^2018-5

=>25^x=4A+5

<=>25^x=(5^2018-5)+5

=>25^x=5^2018

=>25^x=(5^2)^1009=25^1009

=>x=1009

ko bt đúng ko