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Đặt A= 1+3+32+33+...+32009 ta có :
\(3A=3+3^2+3^3+3^4+...+3^{2010}\)
\(3A-A=\left(3+3^2+3^3+3^4+...+3^{2010}\right)-\left(1+3+3^2+3^3+...+3^{2009}\right)\)
\(2A=3^{2010}-1\)
\(A=\frac{3^{2010}-1}{2}\)
A= 1+3+32+33+...+32009
3A=3+3^2+3^3+3^4+...+3^{2010}3A=3+32+33+34+...+32010
3A-A=\left(3+3^2+3^3+3^4+...+3^{2010}\right)-\left(1+3+3^2+3^3+...+3^{2009}\right)3A−A=(3+32+33+34+...+32010)−(1+3+32+33+...+32009)
2A=3^{2010}-12A=32010−1
A=\frac{3^{2010}-1}{2}A=232010−1
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)Đặt A=1+2+22+23+.....+22008
=>2A=2+22+23+....+22009
=>2A-A=(2+22+23+...+22009)-(1+2+22+23+....+22008)
=-1+22009
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
GỌI \(A=1+2^2+2^3+...+2^{2008}\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{2009}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2009}\right)-\left(1+2^2+2^3+...+2^{2009}\right)\)
\(\Rightarrow A=2^{2009}-1\)
ta có:\(2^{2009}-1+1-2^{2009}=0\)
=> A và mẫu số đối nhau
=>\(B=\frac{A}{-1.A}=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(-17.53-21.\left(-17\right)-17.17^0\)
\(=\)\(-17.\left(53-21\right)-17.1\)
\(=\) \(-17.32-17\)
\(=\)\(-544-17\)
\(=-561\)
b, \(\left|-2^2.2^3-3^5\right|+3^5+2009^0-\left(-1\right)^{101}\)
\(=\left|-4.8-243\right|+243+1+1^{101}\)
\(=\left|-32-243\right|+243+1+1\)
\(=\left|-275\right|+243+1+1\)
\(=275+243+1+1\)
\(=518+1+1\)
\(=519+1\)
\(=520\)
:)
3B=3+32+33+...+32010
-B=1+3+32+...+32009
2B=32010-1
B=(32010-1)/2
\(B=1+3+3^2+3^3+...+3^{2009}\)
\(3B=3+3^2+3^3+...+3^{2010}\)
\(3B-B=3^{2010}-1\)
\(2B=3^{2010}-1\)
\(\Rightarrow B=\frac{3^{2010}-1}{2}\)
học tốt