\(\Rightarrow2A=2+2^2+2^3+...+2^{2003}\\ \Rightarrow2A-A=2+2^2+2^3+...+2^{2003}-1-2-...-2^{2002}\\ \Rightarrow A=2^{2003}-1=B\)

\(A=1+2+2^2+...+2^{2002}\)

\(2A=2+2^2+2^3+...+2^{2003}\)

\(2A-A=\left(2+2^2+2^3+...+2^{2003}\right)-\left(1+2+2^2+...+2^{2002}\right)\)

\(A=2^{2003}-1\)

⇒ \(A=B\)

A = \(\frac{2004-2003}{2004+2003}\)và  B = \(\frac{2004^2-2003^2}{2004^2+2003^2}\)

Ta đặt : 2004 = x

             2003 = y

Theo tính chất cơ bản của phân thức , ta có :

\(\frac{x-y}{x+y}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)\left(x+y\right)}=\frac{x^2-y^2}{x^2+y^2+2xy}\)       ( 1 )

Vì x > 0 , y > 0 nên x² + y² + 2xy > x² + y²

\(\Rightarrow\frac{x^2-y^2}{x^2+y^2+2xy}< \frac{x^2-y^2}{x^2+y^2}\)      ( 2 )

Từ ( 1 ) và ( 2 ) 

\(\Rightarrow\frac{x-y}{x+y}< \frac{x^2-y^2}{x^2+y^2}\)

Vậy A < B

https://h.vn/hoi-dap/tim-kiem?q=so+s%C3%A1nh+2+ph%C3%A2n+s%E1%BB%91++A=+2004%5E2003++1+/+2004%5E2004++1++B=2004%5E2002+1/2004%5E2003++1&id=238505

http://pitago.vn/question/so-sanh-a-frac2004-20032004-2003-va-b-2801.html

A = 1 + 2 + 2² + 2³ + ... + 2²⁰⁰²

=> 2A = 2 + 2² + 2³ + 2⁴ + ... + 2²⁰⁰³

=> 2A - A = ( 2 + 2² + 2³ + 2⁴ + ... + 2²⁰⁰³ ) - ( 1 + 2 + 2² + 2³ + ... + 2²⁰⁰² )

A = 2²⁰⁰³ - 1 < 2²⁰⁰³ 

hay A < B

Vậy ...

\(A=1+2+2^2+2^3+...+2^{2002}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{2002}+2^{2003}\)

\(\Rightarrow2A-A=2^{2003}-1\)

\(\Rightarrow A=2^{2003}-1\)

Vì \(2^{2003}-1< 2^{2003}\)

nên A < B

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

A = 1 + 2 + 2² + ... + 2^2002  

A = 1 + (2 + 2² + ... + 2^2002 )  

Ta xét :  

u1 = 2  

u2 = 2.2 = 2²  

u3 = 2.2² = 2^3  

u2002 = 2.2^2001 = 2^2002  

Tổng cấp số nhân : S = u1.(1 - q^n) / (1 - q) = 2.(1 - 2^2002) / (1 - 2) = 2(2^2002 - 1) = 2^2003 - 2  

A = 1 + 2^2003 - 2 = 2^2003 - 1  

So sánh với B  

2^2003 - 1 = 2^2003 - 1

 Vậy B = A 

A<B                      

=>2A=2+2^2+2^3+2^4+2^5+...+2^2002+2^2003

=>2A-A=2^2003-1

=>A=2^2003-1

=>A<B

ta có : a = 1 + 2 + 2^2 + ... + 2^2002

=> 2a = 2 + 2^2 + 2^3 + ... + 2^2003

=> 2a-a = (2+2^2 + 2^3 + ... + 2^2003) - ( 1+2+2^2+...+2^2002)

=> a = 2^2003 - 1

Vậy a=b

Ta có:

\(B=2^{2012}+2^{2011}+...+2^3+2^2+2+1\)

\(\Rightarrow2B=2^{2013}+2^{2012}+...+2^4+2^3+2^2+2\)

\(\Rightarrow2B-B=\left(2^{2013}+2^{2012}+...+2^4+2^3+2^2+2\right)-\left(2^{2012}+...+1\right)\)

\(\Rightarrow B=2^{2013}-1\)

\(A=2^{2003}.9+2^{2003}.1005\)

\(\Rightarrow A=2^{2003}.\left(9+1005\right)\)

\(\Rightarrow A=2^{2003}.1024\)

\(\Rightarrow A=2^{2003}.2^{10}\)

\(\Rightarrow A=2^{2013}\)

Vì \(2^{2013}-1< 2^{2013}\) nên A > B

Vậy A > B