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a)Có A=5+52+53+...+52016
=>5A=52+53+...+52017
=>4A=5A-A=52017-5
=>4A+5=52017-5+5=52017=5x
=>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
\(A=5+5^2+5^3+...+5^{2016}\)
\(5A=5^2+5^3+5^4+...+5^{2017}\)
\(\rightarrow5A-A=5^{2017}-5\)
\(4A=5^{2017}-5\)
\(\Rightarrow4A+5=5^{2017}-5+5\)
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 + 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
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
5x=52018
x=2018
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 = 5n-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 = 5n-1
\(\Rightarrow\) 4A + 5 = 52017 - 5 + 5 = 52017 = 5n-1
\(\Rightarrow\) n - 1 = 2017
\(\Rightarrow\) n = 2018
Vậy n = 2018
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 : \(25^x=4A+5\Leftrightarrow25^x=5^{2018}\)
\(\Leftrightarrow5^{2x}=5^{2018}\Leftrightarrow2x=2018\Leftrightarrow x=2014\)
\(5A=5^2+5^3+5^4+...+5^{2017}\)
\(4A=5A-A=5^{2017}-5\)
\(\Rightarrow4A+5=5^{2017}-5+5=5^{2017}=5^x\Rightarrow x=2017\)