\(A=\left(-7\right)+\left(-7\right)^2+......+\left(-7\right)^{2006}+\left(-7\right)^{2007}\)
\(=\left[\left(-7\right)+\left(-7\right)^2+\left(-7\right)^3\right]+\left[\left(-7\right)^4+\left(-7\right)^5+\left(-7\right)^6\right]+.......\) \(+\left[\left(-7\right)^{2005}+\left(-7\right)^{2006}+\left(-7\right)^{2007}\right]\)
\(=\left(-7\right)\left[1+\left(-7\right)+\left(-7\right)^2\right]+......+\left(-7\right)^{2005}\left[1+\left(-7\right)+\left(-7\right)^2\right]\)
\(=\left(-7\right).43+\left(-7\right)^3.43+......+\left(-7\right)^{2005}.43\)
\(=43\left[\left(-7\right)+\left(-7\right)^3+.....+\left(-7\right)^{2005}\right]\).
Suy ra A chia hết cho 43.

A=(-7+-7^2+-7^3)+.....+(-7^2005+-7^2006+-7^2007)

A=-7(1+-7+-7^2)+.....+-7^2005(1+-7+-7^2)

A=-7.43+....+-7^2005.43\(⋮\)43\(\Rightarrow\)dpcm

b)\(m^2-2mn+n^2+3mn\)

=\(\left(m-n\right)^2+3mn⋮9\)

=\(3mn⋮3\)

\(\Rightarrow\left(m-n\right)^2⋮3\)

\(\Rightarrow\left(m-n\right)^2⋮9\)

\(\Rightarrow3mn⋮9\)

\(\Rightarrow mn⋮3\)

\(\Rightarrow\)m hoạc n\(\)\(⋮\)3

Giả sử m\(⋮\)3,m-n\(⋮\)

\(\Rightarrow\)n\(⋮3\)

\(\Rightarrow\)dpcm

a giải luôn cho e nhé

7A=7+7²+7³+...+7²⁰⁰⁸

7A-A=[7+7²+7³+...+7²⁰⁰⁸]-[1+7+7²+..+7²⁰⁰⁷]

6A=7²⁰⁰⁸-1

A=7²⁰⁰⁸-1/6

b,Tương tư nhân B vs 4 là ra

Mình chỉ trả lời được 2 câu đầu thôi nhé:

a.A= \(1+7+7^2+7^3+...+7^{2007}\)

A.7 = \(7+7^2+7^3+7^4+...+7^{2008}\)

A7-A = \(\left(7+7^2+7^3+7^4+...+7^{2008}\right)-\left(1+7+7^2+7^3+...+7^{2007}\right)\)

A6 =\(7^{2008}-1\)

\(\Rightarrow A=7^{2008}-1\)

Câu còn lại làm tương tự bạn nhé

A=(7²⁰⁰⁸-1):6

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

\(A=\left[\left(-7\right)+\left(-7\right)^2+\left(-7\right)^3\right]+...+\left[\left(-7\right)^{2005}+\left(-7\right)^{2006}+\left(-7\right)^{2007}\right]\)

\(A=-7.\left[1+\left(-7\right)+49\right]+\left(-7\right)^4.\left[1+\left(-7\right)+49\right]+...+\left(-7\right)^{2005}.\left[1+\left(-7\right)+49\right]\)

\(A=-7.43+\left(-7\right)^4.43+...+\left(-7\right)^{2005}.43\)

\(A=43\left[\left(-7\right)+\left(-7\right)^4+...+\left(-7\right)^{2005}\right]⋮43\)

Vậy A chia hết cho 43.

tổng A luôn chia hết nha bạn

A = (-7) + (-7)² + ...+ (-7)²⁰⁰⁶ + (-7)²⁰⁰⁷

A = [ (-7) + (-7)² + (-7)³ ] + [ (-7)⁴ + (-7)⁵ + (-7)⁶ ] + ... + [ (-7)²⁰⁰⁵ + (-7)²⁰⁰⁶ + (-7)²⁰⁰⁷ ]

A = (-7) . [ 1 + (-7) + (-7)² ] + (-7)⁴ . [ 1+ (-7) + (-7)² ] + ... + (-7)²⁰⁰⁵ . [ 1 + (-7) + (-7)² ]

A = (-7) . 43 + (-7)⁴ . 43 + ... + (-7)²⁰⁰⁵ . 43

A = 43 . [ (-7) + (-7)⁴ + ... + (-7)²⁰⁰⁵ ]

=>A chia hết cho 43

Vậy A chia hết cho 43

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

Sửa đề: Tính tổng:

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

Giải:

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

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

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

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

\(\Rightarrow8A=-7+7^{2008}\Rightarrow A=\dfrac{-7+7^{2008}}{8}\)

Vậy \(A=\dfrac{-7+7^{2008}}{8}\)

_____________________________________

Ta có:

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

\(=\left[\left(-7\right)+\left(-7\right)^2+\left(-7\right)^3\right]+...+\left[\left(-7\right)^{2005}+\left(-7\right)^{2006}+\left(-7\right)^{2007}\right]\)

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

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

\(=43.\left[\left(-7\right)+...+\left(-7\right)^{2005}\right]⋮43\) (Đpcm)

đề sai con cuối