Mik nghĩ là C

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Chọn D

D

Đặt N = 2²⁰⁰⁹ + 2²⁰⁰⁸ + 2²⁰⁰⁷ +......+ 2¹ + 2⁰

2N = 2²⁰¹⁰ + 2²⁰⁰⁹ + 2²⁰⁰⁸ +.....+ 2² + 2¹

2N - N = 2²⁰¹⁰ - 2⁰

=> N = 2²⁰¹⁰ - 1

=> M = 2²⁰¹⁰ - (2²⁰¹⁰ - 1)

=> M = 2²⁰¹⁰ - 2²⁰¹⁰ + 1

=> M = 1 

Đặt N=2²⁰⁰⁹+2²⁰⁰⁸+...+1

=>2N=2²⁰¹⁰+2²⁰⁰⁹+...+2

=>2N-N=(2²⁰¹⁰+2²⁰⁰⁹+...+2)-(2²⁰⁰⁹+2²⁰⁰⁸+...+1)

=>N=2²⁰¹⁰-1

Mà M=2²⁰¹⁰-N=2²⁰¹⁰-(2²⁰¹⁰-1)=1

Ác Mộng trả lời đúng rùi. **** thui

Đặt \(N=2^0+2^1+...+2^{2008}+2^{2009}\)

Suy ra: \(M=2^{2010}-N\)

Ta có: \(N=2^0+2^1+...+2^{2008}+2^{2009}\)

\(\Leftrightarrow2N=2+2^2+...+2^{2009}+2^{2010}\)

\(\Leftrightarrow N=2^{2010}-1\)

\(M=2^{2010}-N=2^{2010}-2^{2010}+1=1\)

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: \(\left\{{}\begin{matrix}x_1+x_2=a\\x_1x_2=a-1\end{matrix}\right.\)

\(M=\dfrac{2x_1^2+x_1x_2+2x_2^2}{x_1^2x_2+x_1x_2^2}=\dfrac{2\left(x_1+x_2\right)^2-3x_1x_2}{x_1x_2\left(x_1+x_2\right)}=\dfrac{2a^2-3a+3}{a^2-a}\)

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

\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)

\(M=2^{2010}-2^{2009}-2^{2008}-...-2^1-2^0\)

\(M=2^{2009}\left(2-1\right)-2^{2008}-...-2^1-2^0\)

\(M=2^{2009}-2^{2008}-2^{2007}-...-2^1-2^0\)

\(M=2^{2008}\left(2-1\right)-2^{2007}-...-2^1-2^0\)

\(M=2^{2008}-2^{2007}-2^{2006}-...-2^1-2^0\)

...........................................

\(M=2^1-2^0=2-1=1\)

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

\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)

Gọi \(N=2^{2009}+2^{2008}+...+2^1+2^0\)

\(2N=2^{2010}+2^{2009}+...+2^2+2^1\\ 2N-N=\left(2^{2010}+2^{2009}+...+2^2+2^1\right)-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\\ N=2^{2010}-2^0\\ N=2^{2010}-1\)

Thay vào ta được

\(M=2^{2010}-\left(2^{2010}-1\right)\\ M=2^{2010}-2^{2010}+1\\ M=1\)

Vậy \(M=1\)

Ta có :

\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^0\right)\)

Đặt A=2²⁰⁰⁹+2²⁰⁰⁸+..+2⁰

\(A=2^{2009}+2^{2008}+...+2^0\\ 2A=2^{2010}+2^{2009}+...+2^1\\ \Rightarrow2A-A=A=2^{2010}-2^0\\ \Rightarrow M=2^{2010}-\left(2^{2010}-2^0\right)\\ M=2^{2010}-2^{2010}+1\\ \Rightarrow M=1\)

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