ta có 

\(A=\left(-7\right)+\left(-7\right)^2+\left(-7\right)^3+..\left(-7\right)^{2007}\)

\(\Rightarrow-7A=\left(-7\right)^2+\left(-7\right)^3+..+\left(-7\right)^{2008}\)

Lấy hiệu hai đẳng thức ta có 

\(8A=\left(-7\right)-\left(-7\right)^{2008}\Rightarrow A=-\frac{7+7^{2008}}{8}\)

còn A không chia hết cho 43 nhé

1. \(A=\left(2^{2017}\cdot3+2^{2017}\cdot5\right):2^{2018}\)

\(A=\left[2^{2017}.\left(3+5\right)\right]:\left(2^{2018}\right)\)

\(A=\left[2^{2017}.2^3\right]:\left(2^{2018}\right)\)

\(A=2^{2020}:2^{2018}=2^2=4\)

2. a) 2 + x : 5 = 6

=> x : 5 = 4

=> x = 20

b) 5x(7 + 48:x) = 45

=> x(7 + 48:x) = 9

=> 7x + 48 = 9

=> 7x = -39

=> x = -39/7.

c) Không hiểu đề câu này cho lắm.

3. \(25^{30}=\left(5^2\right)^{30}=5^{60};125^{19}=\left(5^3\right)^{19}=5^{57}\)

Vì 60 > 57 => \(25^{30}>125^{19}\)

4. \(S=1+7^1+...+7^{100}\)

\(\Rightarrow7S=7+7^2+...+7^{101}\)

\(\Rightarrow7S-S=7+7^2+...+7^{101}-1-7-...-7^{100}\)

\(\Rightarrow6S=7^{101}-1\)

\(\Rightarrow S=\frac{7^{101}-1}{6}\)

5. \(Q=1+2+2^2+...+2^{49}\)

\(\Rightarrow2Q=2+2^2+...+2^{50}\)

\(\Rightarrow2Q-Q=2+2^2+...+2^{50}-1-2-...-2^{49}\)

\(\Rightarrow Q=2^{50}-1\)

\(\Rightarrow2^{50}-1+1=2^n\)

\(\Rightarrow2^{50}=2^n\Rightarrow n=50\)

A = 1 . (-7) + (-7) . (-7) + (-7) . \(^{\left(-7\right)^2}\)\(+....+1.\left(-7\right)^{2005}+\left(-7\right).\left(-7\right)^{2005}+\left(-7\right)^2.\left(-7\right)^{2005}\)

\(A=\left(-7\right).\left(1+\left(-7\right)+\left(-7\right)^2\right)+...+\left(-7\right)^{2005}.\left(1+\left(-7\right)+\left(-7\right)^2\right)\)

\(A=\left(-7\right).43+....+\left(-7\right)^{2005}.43\)

\(A=43.\left(\left(-7\right)+.....+\left(-7\right)^{2005}\right)\)chia hết cho 43

Vậy A chia hết cho 43

sao tự nhiên lại có dấu = trong [=7] thế kia