Question

a, Tìm số tự nhiên x sao cho

6x chia hết cho 2x + 1

b, Cho A = 1 + 4 + 4² + 4³ + ... + 4⁵⁹

Chứng tỏ rằng A chia hết cho 7

Answer 1

\(b,A=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+...\left(4^{57}+4^{58}+4^{59}\right)\\ A=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...+4^{57}\left(1+4+4^2\right)\\ A=\left(1+4+4^2\right)\left(1+4^3+...+4^{57}\right)\\ A=21\left(1+4^3+...+4^{57}\right)⋮7\)

Answer 2

a: \(\Leftrightarrow2x+1\in\left\{1;3\right\}\)

hay \(x\in\left\{0;1\right\}\)