Bài 1:

a. $2^{29}< 5^{29}< 5^{39}$

$\Rightarrow A< B$

b.

$B=(3^1+3^2)+(3^3+3^4)+(3^5+3^6)+...+(3^{2009}+3^{2010})$

$=3(1+3)+3^3(1+3)+3^5(1+3)+...+3^{2009}(1+3)$

$=(1+3)(3+3^3+3^5+...+3^{2009})$

$=4(3+3^3+3^5+...+3^{2009})\vdots 4$

Mặt khác:

$B=(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2008}+3^{2009}+3^{2010})$

$=3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2008}(1+3+3^2)$

$=(1+3+3^2)(3+3^4+....+3^{2008})=13(3+3^4+...+3^{2008})\vdots 13$

Bài 1:
c.

$A=1-3+3^2-3^3+3^4-...+3^{98}-3^{99}+3^{100}$

$3A=3-3^2+3^3-3^4+3^5-...+3^{99}-3^{100}+3^{101}$

$\Rightarrow A+3A=3^{101}+1$
$\Rightarrow 4A=3^{101}+1$

$\Rightarrow A=\frac{3^{101}+1}{4}$

Bài 2:

a. $7\vdots n+1$

$\Rightarrow n+1\in \left\{1; -1; 7; -7\right\}$

$\Rightarrow n\in \left\{0; -2; 6; -8\right\}$

b.

$2n+5\vdots n+1$
$\Rightarrow 2(n+1)+3\vdots n+1$

$\Rightarrow 3\vdots n+1$

$\Rightarrow n+1\in \left\{1; -1; 3; -3\right\}$

$\Rightarrow n\in \left\{0; -2; 2; -4\right\}$

Xem lại đề có thiếu câu hỏi không nha bạn

Xét hiệu:

a² + b² + c² - ab - bc - ca

=  1 2 (2a² + 2b² + 2c² - 2ab - 2bc - 2ca)

=  1 2 [(a² - 2ab + b²) + (b² - 2bc + c²) + (c² - 2ca + a²)]

= 1 2 [(a - b)² + (b - c)² + (c - a)²] ≥ 0

(vì (a - b)² ≥ 0; (b - c)² ≥ 0; (c - a)² ≥ 0 với mọi a, b, c)

Nên a² + b² + c² ≥ ab + bc + ca.

Dấu “=” xảy ra khi a = b = c.

Đáp án cần chọn là: B

1 

a=(-5;5;-6;6;-7;7;-8;8;-9;9;.......)

2 bằng nhau

1. a thuộ̣c (+-5, +-6, +-7, +-8,..)

2. |a|<|b|

lộn

2 |a|<|b|