a: \(=5^{27}+5^{28}-5^{26}\)

\(=5^{26}\left(5+5^2-1\right)=5^{24}\cdot725⋮725\)

b: \(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{117}\left(1+3+3^2\right)\)

\(=13\left(1+3^3+...+3^{117}\right)\)

\(=13\cdot\left[\left(1+3^3\right)+3^6\left(1+3^3\right)+...+3^{114}\left(1+3^3\right)\right]\)

\(=13\cdot28\cdot\left(1+3^6+...+3^{114}\right)⋮91\)

\(A=7\left(1+7+7^2\right)+7^4\left(1+7+7^2\right)+...+7^{118}\left(1+7+7^2\right)\)

\(=57\left(7+7^4+...+7^{118}\right)⋮57\)

\(A=7\left(1+7+7^2\right)+...+7^{118}\left(1+7+7^2\right)\)

\(=57\left(7+...+7^{118}\right)⋮57\)

Nhiều thế không ai làm đâu bạn          

nhiều nhỉ lấy ở đâu đấy !!!!!!!!!!!!!!!!!!!!!!!!!

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THONG CAM MINH MOI \(y=\frac{1}{x^2+\sqrt{x}}\)7 TUOI

\(91=7.13\)

Đặt \(A=5^n\left(5^n+1\right)-6^n\left(3^{n+2}\right)\)

\(\Rightarrow A=\left(25^n-18^n\right)-\left(12^n-5^n\right)\)

Ta có:

\(\left\{\begin{matrix}25^n-18^n⋮25-18=7\\12^n-5^n⋮12-5=7\end{matrix}\right.\)\(\Leftrightarrow A⋮7\)

Mặt khác:

\(A=\left(25^n-12^n\right)-\left(18^n-5^n\right)\)

Lại có:

\(\left\{\begin{matrix}25^n-12^n⋮25-12=13\\18^n-5^n⋮18-5=13\end{matrix}\right.\)\(\Leftrightarrow A⋮13\)

Mà: \(\left(7;13\right)=1\)

\(\Leftrightarrow A⋮91\)

Vậy \(5^n\left(5^n+1\right)-6^n\left(3^{n+2}\right)⋮91\) (Đpcm)

+) Vì \(3⋮3\); \(3^2⋮3\); \(3^3⋮3\); \(3^4⋮3\); .............. ; \(3^{119}⋮3\); \(3^{120}⋮3\)

\(\Rightarrow3+3^2+3^3+3^4+.........+3^{119}+3^{120}⋮3\)

hay \(A⋮3\)

+) \(A=3+3^2+3^3+3^4+..........+3^{119}+3^{120}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+..........+\left(3^{119}+3^{120}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+.........+3^{119}\left(1+3\right)\)

\(=3.4+3^3.4+........+3^{119}.4=4.\left(3+3^3+.......+3^{119}\right)⋮4\)

+) \(A=3+3^2+3^3+3^4+...........+3^{119}+3^{120}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+........+\left(3^{118}+3^{119}+3^{120}\right)\)

\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+..........+3^{118}\left(1+3+3^2\right)\)

\(=3.13+3^4.13+.......+3^{118}.13=13.\left(3+3^4+........+3^{118}\right)⋮13\)

Vậy \(A⋮3,4,13\)

A = 3 + 3² + 3³ + ... + 3¹²⁰

= 3 (1 + 3 + 3² + ... + 3¹¹⁹) 

Vì 3 chia hết cho 3 nên 3 (1 + 3 + 3² + ... + 3¹¹⁹) chia hết cho 3

=> A chia hết cho 3   (đpcm)

A = 3 + 3² + 3³ + ... + 3¹²⁰

= (3 + 3²) + (3³ + 3⁴) + ... + (3¹¹⁹ + 3¹²⁰)

= 3 (1 + 3) + 3³ (1 + 3) + ... + 3¹¹⁹ (1 + 3)

= 3 . 4 + 3³ . 4 + ... + 3¹¹⁹ . 4

Vì 4 chia hết cho 4 nên 3 . 4 + 3³ . 4 + ... + 3¹¹⁹ . 4 chia hết cho 4

=> A chia hết cho 4   (đpcm)

A = 3 + 3² + 3³ + ... + 3¹²⁰

= (3 + 3² + 3³) + (3⁴ + 3⁵ + 3⁶) + ... + (3¹¹⁸ + 3¹¹⁹ + 3¹²⁰)

= 3 (1 + 3 + 3²) + 3⁴ (1 + 3 + 3²) + ... + 3¹¹⁸ (1 + 3 + 3²)

= 3 . 13 + 3⁴ . 13 + ... + 3¹¹⁸ . 13

Vì 13 chia hết cho 13 nên 3 . 13 + 3⁴ . 13 + ... + 3¹¹⁸ . 13 chia hết cho 13

=> A chia hết cho 13   (đpcm)