1. Cho số nguyên x là 9 (Thỏa mãn x:7, dư 2); 2x+3(giả thuyết)

=> (2.9)+3 = 21 chia hết cho7 (chia hết cho viết bằng ki hiệu nha bạn)

2. 2^0+2^1+2^2+2^3+...+2^5n-3+2^5n-2+2^5-1

= (2^0+2^1+2^2+2^3+2^4)+...+(2^5n-5+2^5n-4+2^5n-3+2^5n-2+2^5n-1)

=(1+2+4+8+16)+...+(2^5n-5+2^5n-4+2^5n-3+2^5n-2+2^5n-1) chia hết cho 31

\(3^{5n+2}+3^{5n+1}-3^{5n}=3^{5n}\left(3^2+3-1\right)=11.3^{5n}⋮11\)

\(3^{5n+2}+3^{5n+1}-3^{5n}(n\in N^*)\\=3^{5n}\cdot3^2+3^{5n}\cdot3-3^{5n}\\=3^{5n}\cdot(3^2+3-1)\\=3^{5n}\cdot11\)

Vì \(3^{5n}\cdot11\vdots11\) 

nên biểu thức \(3^{5n+2}+3^{5n+1}-3^{5n}\vdots11\)

Đặt A = 2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵ + ..... +2^{5n-6 +} 2^5n-5 + 2^5n-4 + 2^5n-3 + 2^5n-2 + 2^5n-1

=> A = ( 2⁰ + 2¹ + 2² + 2³ + 2⁴ + 2⁵ ) + ..... + ( 2^5n-6 + 2^5n-5 + 2^5n-4 + 2^5n-3 + 2^5n-2 + 2^{5n-1 )}

=> A = 2⁰ ( 1 + 2¹ + 2² + 2³ + 2⁴ ) + ..... + 2^5n-6 ( 1 + 2¹ + 2² + 2³ + 2⁴ )

=> A = 1.31 + 2⁵ .31 + ..... + 2^5n-6.31 

=> A = 31.( 1 + 2⁵ + ..... + 2^5n-6 )

Vì 31 ⋮ 31 => A ⋮ 31 ( đpcm )

\(\text{Đặt }A=\left(2^0+2^1+2^2+2^3+2^4\right)+...+\left(2^{5n-5}+2^{5n-4}+2^{5n-3}+2^{5n-2}+2^{5n-1}\right)\)

\(=\left(1+2 +4+8+16\right)+...+2^{5n-5}.\left(2^0+2^1+2^2+2^3+2^4\right)\)

\(=31+...+2^{5n-5}.31\)

\(=31.\left(1+...+2^{5n-5}\right)\text{chia hết cho 31}\left(đpcm\right)\)

5n+5n.52=650

5n(1+52)=650

5n.26=650

=>5n=650:26

=>5n=25=52

=>n=2

3⁵ⁿ⁺²+3⁵ⁿ⁺¹-3⁵ⁿ

=3⁵ⁿ.3²+3⁵ⁿ.3¹-3⁵ⁿ

=3⁵ⁿ.9+3⁵ⁿ.3-3⁵ⁿ

=3⁵ⁿ.(9+3-1)

=3⁵ⁿ.11 chia hết cho 11

=> 3⁵ⁿ⁺²+3⁵ⁿ⁺¹-3⁵ⁿ chia hết cho 11