khỏi ghi lại đề nha 

= 2003 x [ ( 200 + 2004 - 48 ) - ( 52 + 2004 + 89 ) ] 

= 2003 x ( 200 + 2004 - 48 - 52 - 2004 - 89 )
= 2003 x ( 200 - 48 - 52 + 2004 - 2004 - 89 ) 

= 2003 x ( 200 - 100       +          0        - 89 )

= 2003 x ( 100 - 89 ) 

= 2003 x 11

= 22033

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 = \(\dfrac{2004-2003}{2003+2004}\) = \(\dfrac{\left(2004-2003\right).\left(2004+2003\right)}{\left(2003+2004\right).\left(2004+2003\right)}\) =\(\dfrac{2004^2-2003^2}{\left(2003+2004\right)^2}\)

Vì 2003² + 2004² < (2003 + 2004)²

Nên A < B 

ko bit

Ko biết

2003x2005+\(\frac{200}{2004}\)x2004+199

=2003x2005+\(\frac{200x2004}{2004}\)+199

=4016015+200+199

=4016414

\(\dfrac{a+2003}{a-2003}=\dfrac{b-2004}{b+2004}\)

\(\Leftrightarrow\left(a+2003\right)\left(b+2004\right)=\left(a-2003\right)\left(b-2004\right)\)

\(\Leftrightarrow ab+2004a+2003a+2003\cdot2004=ab-2004a-2003a+2003\cdot2004\)

\(\Leftrightarrow4008a=4006b\)

=>a/b=2003/2004

hay a/2003=b/2004