Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
\(A=\frac{2003\times2004-1}{2003\times2004}=\frac{2003\times2004}{2003\times2004}-\frac{1}{2003\times2004}=1-\frac{1}{2003\times2004}\)
\(B=\frac{2004\times2005-1}{2004\times2005}=\frac{2004\times2005}{2004\times2005}-\frac{1}{2004\times2005}=1-\frac{1}{2004\times2005}\)
Vì \(\frac{1}{2003\times2004}>\frac{1}{2004\times2005}\Rightarrow A< B\)
1)2003x2005+200 2)2004x2004+199
=(200x10+3)x2005+200 =2004x2004+(200-1)
=200x10x2005+3x2005+200 =2004x2004+200-1
=200x(10x2005+1)+3x2005 =(200x10+4)x2004+200-1
=200x20051+6015 =200x10x2004+4x2004+200-1
=4010200+6015 =200x(10x2004+1)+4x2004-1
=4016215 =200x20041+8016-1
=4008200+8015 =4016215
chọn mình nha
\(\frac{2004\times13+1990+2002\times2003}{2003+2003\times501+506\times2003}=\frac{2004\times13+1990+2002}{\left(2003+1\right)\times501+506}=\frac{2004\times4205}{2004\times1007}=\frac{4205}{1007}\)
1-2-3+4+5-6-7+8+...+197-198-199+200
=(1-2-3+4)+(5-6-7+8)+...+(197-198-199+200)
=0+0+...+0
=0
\(=2003\times\left(58+52-10\right)+335\times100\)
\(=100\times\left(2003+335\right)=233800\)
(2003 x 58 + 52 x 2003 - 10 x 2003) + (670-335) x 100
= 2003 x ( 58 + 52 - 10 ) + 335 x 100
= 2003 x 100 + 335 x 100
= ( 2003 + 335 ) x 100
= 2338 x 100
233 800
y=\(\frac{2006x2005-1}{2004x2006+2005}=\frac{2006x2005-1}{\left(2005-1\right)x2006+2005}=\frac{2006x2005-1}{2005x2006-2006+2005}=\frac{2006x2005-1}{2005x2006-1}=1\)
(2004-1)(2004+1)+200/2004.2004+199
=2004^2-1+200/2004^2+199
=2004^2+199/2004^2+199
=1
\(\frac{2003.2005+200}{2004.2004+199}\)
= \(\frac{2003.\left(2004+1\right)+200}{\left(2003+1\right).2004+199}\)
= \(\frac{2003.2004+2003+200}{2003.2004+2004+199}\)
= \(\frac{2003.2004+2203}{2003.2004+2203}\)
= \(1\)
~~~~Hok tốt~~~~