b) \(A=1+5+5^1+5^2+5^3+...+5^{71}\)

\(\Rightarrow A=\left(1+5^1+5^2\right)+5^3\left(1+5^1+5^2\right)+...+5^{69}\left(1+5^1+5^2\right)\)

\(\Rightarrow A=31+5^3.31+...+5^{69}.31\)

\(\Rightarrow A=31\left(1+5^3+...+5^{69}\right)⋮31\left(dpcm\right)\)

a) \(A=1+5^1+5^2+5^3+...+5^{71}\)

\(\Rightarrow A=\dfrac{5^{71+1}-1}{5-1}=\dfrac{5^{72}-1}{4}\)

\(4A+x=5^{72}\)

\(\Rightarrow4.\dfrac{5^{72}-1}{4}+x=5^{72}\)

\(\Rightarrow5^{72}-1+x=5^{72}\)

\(\Rightarrow x=1\)

bài 1 có ý d nha các bạn mình viết thiếu

Bài dái quá, bạn nên tách ra đi nhé!

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 ..

Sửa đề: \(A=2+2^2+2^3+\cdots+2^{10}+1+5+5^2+\cdots+5^{14}\)

a: Đặt \(B=2+2^2+2^3+\cdots+2^{10}\)

\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\)

\(=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)\)

\(=31\cdot\left(2+2^6\right)\) ⋮31

Đặt \(C=1+5+5^2+\cdots+5^{14}\)

\(=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+\cdots+\left(5^{12}+5^{13}+5^{14}\right)\)

\(=\left(1+5+5^2\right)+5^3\left(1+5+5^2\right)+\cdots+5^{12}\left(1+5+5^2\right)\)

\(=31\left(1+5^3+\cdots+5^{12}\right)\) ⋮31

Ta có: \(A=2+2^2+2^3+\cdots+2^{10}+1+5+5^2+\cdots+5^{14}\)

=B+C

mà B⋮31 và C⋮31

nên A⋮31

b: Ta có: \(B=2+2^2+2^3+\cdots+2^{10}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\cdots+\left(2^9+2^{10}\right)\)

\(=\left(2+2^2\right)+2^2\left(2+2^2\right)+\cdots+2^8\left(2+2^2\right)\)

\(=6\left(1+2^2+\cdots+2^8\right)\) ⋮6

Ta có: \(C=1+5+5^2+\cdots+5^{14}\)

\(=1+\left(5+5^2\right)+\left(5^3+5^4\right)+\cdots+\left(5^{13}+5^{14}\right)\)

\(=1+5\left(1+5\right)+5^3\left(1+5\right)+\cdots+5^{13}\left(1+5\right)\)

\(=1+6\left(5+5^3+\cdots+5^{13}\right)\)

=>C chia 6 dư 1

Ta có: A=B+C

\(=6\left(1+2^2+\cdots+2^8\right)+1+6\left(5+5^3+\cdots+5^{13}\right)\)

=>A chia 6 dư 1

12 + 14 + 16 + x chia hết cho 2

12 ; 14 ; 16 chia hết cho 2 => x chia hết cho 2

12 + 14 + 16 không chia hết cho 2

12 ; 14 ; 16 chia hết cho 2 => x không chia hết cho 2 (lẻ)   

1.A=5+5²+....+5¹⁰⁰

<=> 5A=5²+5³+.....+5¹⁰¹

<=> 5A-A=(5²+5³+....+5¹⁰¹)-(5+5²+....+5¹⁰⁰)

<=> 4A=5¹⁰¹-5

<=> \(A=\frac{5^{101}-5}{4}\)

2. Ta có : 4A=5¹⁰¹-5

<=> 4A+5=5¹⁰¹

Vậy x=101.

3. \(A=5+5^2+....+5^{100}\)

\(\Rightarrow A=\left(5+5^2+5^3+5^4\right)+...+\left(5^{97}+5^{98}+5^{99}+5^{100}\right)\)

\(\Rightarrow A=5.\left(1+5+25+125\right)+...+5^{97}.\left(1+5+25+125\right)\)

\(\Rightarrow A=5.165+....+5^{97}.165\)

\(\Rightarrow A=165.\left(5+...+5^{97}\right)\)

\(\Rightarrowđpcm\)

Mình xin lỗi viết nhầm 

\(A=156.\left(5+....+5^{97}\right)\)

\(\Rightarrow A⋮156\)

giup minh voi sap phai nop roi

câu a Achia hết cho 128

4A+1 là số chính phương

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