Vậy ta thấy 5A=5+5^2+5^3+5^4+...+5^2009+5^2010

=> 5A-A= 5^2010-1

=> 4A=5^2010-1=> 4A=(5^2010-1)/4

 đến đaay em tính ra bằng máy tính hay để nguyên thì chắc chắn cô giáo sẽ cho điểm, tốt nhất cứ để nguyên nhé :)

Nguyễn đức hiếu làm sai kìa 

Đoạn cuối :

4A = 5²⁰²⁰ -1 

\(A = { {5mũ2020-1} \over 4}\)

\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)

\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)

\(=\frac{1}{5}+\frac{2}{3}\)

\(=\frac{13}{15}\)

Bài làm:

\(A=1-2+3-4+5-...-2008+2009\)

\(A=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(2007-2008\right)+2009\)

\(A=-1-1-1-...-1+2009\)(1004 số -1)

\(A=-1004+2009=1005\)

\(B=1+2-3-4+5+6-7-...-2007-2008+2009+2010\)

\(B=1+\left(2-3-4+5\right)+\left(6-7-8+9\right)+...+\left(2006-2007-2008+2009\right)+2010\)

\(B=1+0+0+...+0+2010\)

\(B=2011\)

Học tốt!!!!

nhân 5 lần lên:

5A=5+5²+...+5²⁰¹⁰

=> 4A =5A-A= 5²⁰¹⁰-1  => A= (5²⁰¹⁰-1):4

      5A = \(5+5^2+5^3+5^4+...+5^{2009}+5^{2010}\)

        A = \(1+5+5^2+5^3+...+5^{2008}+5^{2009}\)

\(\Rightarrow\) 4A = \(5^{2010}-1\)

\(\Rightarrow\)   A = \(\frac{5^{2010}-1}{4}\)

Đúng thì cho mk biết nha

5B = 5 +5² +5³+....+5²⁰⁰⁹

5B- B  = 4B= (5²⁰¹⁰-1)

=> B= \(\frac{5^{2010}-1}{4}\)

**** CHO MÌNH NHA 

\(A=1+5+5^2+5^3+...+5^{2008}+5^{2009}\)

\(\Rightarrow5A=5+5^2+5^3+5^4+...+5^{2009}+5^{2010}\)

\(\Rightarrow5A-A=\left(5+5^2+5^3+5^4+...+5^{2009}+5^{2010}\right)-\left(1+5+5^2+5^3+....+5^{2008}+5^{2009}\right)\)

\(\Rightarrow4A=5^{2010}-1\)

\(\Rightarrow A=\frac{5^{2010}-1}{4}\)

A=1+5+5²+...+5²⁰⁰⁹

=>5A=5+5²+5³+...+5²⁰¹⁰

=>4A=5²⁰¹⁰-1

=>A=5²⁰¹⁰-1/4

\(B=1+5+5^2+5^3+...+5^{2008}+5^{2009}\)
Đặt \(5B=5.\text{ (}1+5+5^2+5^3+...+5^{2008}+5^{2009}\text{)}\)
\(\Rightarrow5B=5+5^2+5^3+...+5^{2009}+5^{2010}\)
\(\Rightarrow5B-B=\left(5+5^2+5^3+...+5^{2009}+5^{2010}\right)-\left(1+5+5^2+...+5^{2009}\right)\)\(\Rightarrow4B=5^{2010}-1\)
\(\Rightarrow B=\dfrac{5^{2010}-1}{4}\)
Vậy \(B=\dfrac{5^{2010}-1}{4}\)