tính:
\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=2^{2010}-2^{2009}-2^{2008}-...-2^1-2^0\)
\(-M=-\left(2^{2010}-2^{2009}-2^{2008}-...-2^1-2^0\right)\)
\(-M=2^{2010}+2^{2009}+2^{2008}+...+2^1+2^0\)
\(-2M=2.\left(2^{2010}+2^{2009}+2^{2008}+...+2^1+2^0\right)\)
\(-2M=2^{2011}+2^{2010}+2^{2009}+...+2^2+2^1\)
\(-M=2^{2011}+2^{2010}+...+2^2+2^1-\left(2^{2010}+2^{2009}+2^{2008}+...+2^1+2^0\right)\)
\(-M=2^{2011}-1=>M=-2^{2011}+1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
Vì a,b là nghiệm PT nên \(\left\{{}\begin{matrix}30a^2-4a=2010\\30b^2-4b=2010\end{matrix}\right.\)
\(\Rightarrow N=\dfrac{a^{2008}\left(30a^2-4a\right)+b^{2008}\left(30b^2-4b\right)}{a^{2008}+b^{2008}}\\ \Rightarrow N=\dfrac{a^{2008}\cdot2010+b^{2008}\cdot2010}{a^{2008}+b^{2008}}=2010\)
Bài 1:
Viét: \(\left\{{}\begin{matrix}x_1+x_2=a\\x_1x_2=a-1\end{matrix}\right.\)
\(M=\dfrac{2x_1^2+x_1x_2+2x_2^2}{x_1^2x_2+x_1x_2^2}=\dfrac{2\left(x_1+x_2\right)^2-3x_1x_2}{x_1x_2\left(x_1+x_2\right)}=\dfrac{2a^2-3a+3}{a^2-a}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A=\(\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)
=> 2A=\(2^{2010}+2^{2009}+2^{2008}+...+2^2+2\)
=>2A-A=\(\left(2^{2010}+2^{2009}+2^{2008}+...+2^2+2\right)-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)
=> A=22010-20
=>M=22010-A=22010-22010+20=1
đúng nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
xin hỏi nguyên:tại sao bạn lại viết được hai chuyên mục cùng một lúc được(lũy thừa và tính nhanh)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=2^{2009}+2^{2008}+...+2+2^0\)
\(=1+2+...+2^{2008}+2^{2009}\)
\(\Rightarrow2A=2+2^2+...+2^{2010}\)
\(\Rightarrow2A-A=\left(2+2^2+...+2^{2010}\right)-\left(1+2+...+2^{2009}\right)\)
\(\Rightarrow A=2^{2010}-1\)
\(\Rightarrow M=2^{2010}-\left(2^{2010}-1\right)\)
\(=2^{2010}-2^{2010}+1=1\)
Vậy M = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)
\(M=2^{2010}-2^{2009}-2^{2008}-...-2^1-2^0\)
\(M=2^{2009}\left(2-1\right)-2^{2008}-...-2^1-2^0\)
\(M=2^{2009}-2^{2008}-2^{2007}-...-2^1-2^0\)
\(M=2^{2008}\left(2-1\right)-2^{2007}-...-2^1-2^0\)
\(M=2^{2008}-2^{2007}-2^{2006}-...-2^1-2^0\)
...........................................
\(M=2^1-2^0=2-1=1\)
đặt M1 = 22009 + 22008 +...+21 + 20
⇒ 2M1 = 22010 + 22009 + ... + 22 + 21
⇒ 2M1 - M1 = 22010 + 22009 + ... + 22 + 21 - (22009 + 22008 + ... + 21 + 20)
⇒ M1 = 22010 - 20
⇒ M = 22010 - (22010 - 20)
⇒ M = 22010 - 22010 +20
⇒ M = 0 + 1 = 1
Vậy M = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=2^{2009}+2^{2008}+...+2^1+2^0.\)
Ta có : \(2A=2^{2010}+2^{2009}+...+2^2+2^1.\)
Suy ra : \(2A-A=2^{2010}-2^0\Rightarrow A=2^{2010}-1.\)
Do đó \(M=2^{2010}-A=2^{2010}-\left(2^{2010}-1\right)=1.\)
Đặt A=22009+22008+...+21+20.A=22009+22008+...+21+20.
Ta có : 2A=22010+22009+...+22+21.2A=22010+22009+...+22+21.
Suy ra : 2A−A=22010−20⇒A=22010−1.2A−A=22010−20⇒A=22010−1.
Do đó M=22010−A=22010−(22010−1)=1.
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=2^{2009}+2^{2008}+2^{2007}+...+2+1\\ \Rightarrow2A=2^{2010}+2^{2009}+2^{2008}+...+2^2+2\\ \Rightarrow2A-A=\left(2^{2010}+2^{2009}+2^{2008}+...+2^2+2\right)-\left(2^{2009}+2^{2008}+2^{2007}+...+2+1\right)\\ \Rightarrow A=2^{2010}-1\)
\(\Rightarrow M=2^{2010}-2^{2010}+1=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=2^{2009}+2^{2008}+...+2^1+2^0\)
Ta có: \(2A=2^{2010}+2^{2008}+...+2^1\)
=> \(2AtrừA=\left(2^{2010}+2^{2008}+...+2^1\right)trừ\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)
=> \(A=2^{2010}trừ1\)
Thay vào ta có:
\(M=2^{2010}trừ2^{2010}trừ1\)
\(\Rightarrow M=âm1\)
Xin lỗi. Máy mình không có dấu trừ. Nên viết thành chữ.
Đặt \(A=2^{2009}+2^{2008}+...+2^1+2^0\)
Ta có : \(2A=2^{2010}+2^{2009}+...+2^2+2^1\)
\(\Rightarrow2A-A=2^{2010}-2^0\Rightarrow A=2^{2010}-1\)
Do đó : \(M=2^{2010}-A=2^{2010}-\left[2^{2010}-1\right]=1\)
\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)
\(2^{2010}-M=2^{2009}+2^{2008}+...+2+1\)
\(2\left(2^{2010}-M\right)=2\left(2^{2009}+2^{2008}+...+2+1\right)\)
\(2\left(2^{2010}-M\right)=2^{2010}+2^{2009}+...+2^2+2\)
\(2\left(2^{2010}-M\right)-M=\left(2^{2010}+2^{2009}+...+4+2\right)-\left(2^{2009}+2^{2008}+...+2+1\right)\)
\(2^{2010}-M=2^{2010}+2^{2009}+...+4+2-2^{2009}-2^{2008}-...-2-1\)
\(2^{2010}-M=2^{2010}-1\)
=> M = 1