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

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

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

=> \(8A=-7-7^{2008}\) => \(A=-\frac{7+7^{2008}}{8}\)

b) \(A=\left(\left(-7\right)+\left(-7\right)^2+\left(-7\right)^3\right)+...+\left(\left(-7\right)^{2005}+\left(-7\right)^{2006}+\left(-7\right)^{2007}\right)\) ( Chia thành 2007 : 3 = 669 nhóm 3 số)

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

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

Vậy A chia hết cho 43

A= (- 7) + (-7)^2+ … + (- 7)^2006 + (- 7)^2007 

<=> -7A = (-7)^2+ … + (- 7)^2006 + (- 7)^2008 

A-(- 7A )= (- 7) + (-7)^2+ … + (- 7)^2006 + (- 7)^2007-{(-7)^2+ … + (- 7)^2006 + (- 7)^2008} 

<=> 8A = -7 - (- 7)^2008 = -7 + 7^2008 = 7^2008 - 7 

<=> A = (7^2008 - 7)/8 .

Cảm ơn cô Loan cô xn ycầu kb cho e nhá !!!! Ths cô nhìu...

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

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

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

7A-A=6A= 7²⁰⁰⁸- 7

=> A= \(\frac{7^{2008}-7}{6}\)

Mình làm vậy ko biết có đúng ko nữa

Nguyễn Huy Thắng, Nguyễn Huy Tú

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

Khó thế!

a)  \(\Leftrightarrow\frac{x+7}{2003}+1+\frac{x+4}{2006}+1-\frac{x-1}{2011}-1-\frac{x-5}{2015}-1=0\)

     \(\Leftrightarrow\frac{x+2010}{2003}+\frac{x+2010}{2006}-\frac{x+2010}{2011}-\frac{x+2010}{2015}=0\)

     \(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2003}+\frac{1}{2006}-\frac{1}{2011}-\frac{1}{2015}\right)=0\)

     \(\Leftrightarrow x+2010=0\) ( vì 1/2003  +  1/2006  --  1/2011  -- 1/2015   \(\ne\)0)

    \(\Leftrightarrow x=-2010\)

câu b làm tương tự (có gì không hiểu hỏi mk nha) >v<

a)\(10\left(x-7\right)-8\left(x+5\right)=6\cdot\left(-5\right)+24\)

\(10x-10\cdot7-8x-8\cdot5=\left(-30\right)+24\)

\(10x-70-8x-40=-6\)

\(10x-8x=\left(-6\right)+70+40\)

\(2x=104\)

\(x=104\div2\)

\(x=52\)

b)\(2\left(4x-8\right)-7\left(3+x\right)=6\)

\(2\cdot4x-2\cdot8-7\cdot3-7x=6\)

\(8x-16-21-7x=6\)

\(8x-7x=6+16+21\)

\(x=43\)