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

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

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

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

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

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

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

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

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

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

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

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

Đặt \(A=2^{2009}+2^{2008}+2^{2007}+...+2+1\\ \Rightarrow2A=2^{2010}+2^{2009}+2^{2008}+...+2^2+2\\ \Rightarrow2A-A=\left(2^{2010}+2^{2009}+2^{2008}+...+2^2+2\right)-\left(2^{2009}+2^{2008}+2^{2007}+...+2+1\right)\\ \Rightarrow A=2^{2010}-1\)

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

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

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

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

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

\(-2M=2^{2011}+2^{2010}+2^{2009}+...+2^2+2^1\)

\(-M=2^{2011}+2^{2010}+...+2^2+2^1-\left(2^{2010}+2^{2009}+2^{2008}+...+2^1+2^0\right)\)

\(-M=2^{2011}-1=>M=-2^{2011}+1\)

tại sao lại có dấu ''-'' vậy bạn mình không hiểu lắm.

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!

xin hỏi nguyên:tại sao bạn lại viết được hai chuyên mục cùng một lúc được(lũy thừa và tính nhanh)

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

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

\(2S=2^1+2^2+...+2^{2009}+2^{2010}\)

\(2S-S=S=2^{2010}-2^0\)

Thay S vào M ta được: \(M=2^{2010}-\left(2^{2010}-2^0\right)=2^{2010}-2^{2010}+2^0=1\)

Vậy \(M=1\)

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

Ta có :

Suy ra :

Do đó

Bạn dựa theo bài của Việt mà làm

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.