1) \(5+5^2+5^3+.....+5^{12}=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{11}+5^{12}\right)\)

\(=30.1+5^2.30+.....+5^{10}.30=30.\left(1+5^2+....+5^{10}\right)\)

Vậy chia hết cho 30

\(5+5^2+5^3+....+5^{12}=\left(5+5^2+5^3\right)+.....+\left(5^{10}+5^{11}+5^{12}\right)\)

\(=5.31+5^4.31+....+5^{10}.31=31.\left(5+5^4+....+5^{10}\right)\)

Vậy chia hết cho 31

haizzzzzzzzzzz câu 2 làm tek nào z

Bài làm :

Ta thấy : 4 + 3 = Tứ + Tam 

Tứ + Tam = Tám + Tư

Mà Tám + Tư = 8 + 4 = 12

=> 4 + 3 = 12

Mình làm đúng rùi đó !k mình nha!

tu cong tam = tam cong tu

ma 8+4=12

tứ + tam = tám + tư = 12

kb nha

a) \(A=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+...+\frac{1}{60}\right)+...+\frac{1}{70}\)

Nhận xét:

\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\ge\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)

\(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{30}\ge\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)

\(\frac{1}{31}+...+\frac{1}{60}\ge\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{30}{60}=\frac{1}{2}\)

\(A\ge\frac{1}{2}+\frac{1}{3}+\frac{1}{2}+\frac{1}{61}...+\frac{1}{70}\ge\frac{1}{2}+\frac{1}{3}+\frac{1}{2}=\frac{4}{3}\)

Sorry ,tất cả dấu lớn hơn hoặc bằng đổi thành dấu > nhé 

còn câu b

https://olm.vn/hoi-dap/detail/54833154236.html

Câu 2:

\(C=3^{10}+3^{11}+3^{12}+...+3^{17}.\)

\(C=\left(3^{10}+3^{11}+3^{12}+3^{13}\right)+\left(3^{14}+3^{15}+3^{16}+3^{17}\right).\)

\(C=3^{10}\left(1+3+3^2+3^3\right)+3^{14}\left(1+3+3^2+3^3\right).\)

\(C=3^{10}\left(1+3+9+27\right)+3^{14}\left(1+3+9+27\right).\)

\(C=3^{10}.40+3^{14}.40.\)

\(C=\left(3^{10}+3^{14}\right).40⋮40\left(đpcm\right).\)

\(C=3^{10}+3^{11}+..+3^{17}\\ =\left(3^{10}+3^{11}+3^{12}+3^{13}\right)+\left(3^{14}+..+3^{17}\right)\\ =3^{10}\left(1+3+3^2+3^3\right)+3^{14}\left(1+3+3^2+3^3\right)\\ =40\left(3^{10}+3^{14}\right)⋮40\)

1)

+Nếu x lẻ thì x+2017 là chẵn \(⋮2\)

+Nếu x là chẵn thì x+2018 cũng là chãn \(⋮2\)

\(\Rightarrow dpcm\)

Chỉnh đề:

Ta có:

\(A=2+2^2+2^3+2^4+...2^{12}\)

\(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{10}+2^{11}+2^{12}\right)\)

\(A=14+2^3.\left(2+2^2+2^3\right)+...+2^9.\left(2+2^2+2^3\right)\)

\(A=14+2^3.14+...+2^9.14\)

\(A=14.\left(1+2^3+...+2^9\right)\)

Vì \(14⋮7\) nên \(14.\left(1+2^3+...2^9\right)⋮7\)

Vậy \(A⋮7\)