\(\dfrac{2004.2005-1}{2004.2005}=1-\dfrac{1}{2004.2005}\)

\(\dfrac{2005.2006-1}{2004.2006}=1-\dfrac{1}{2005.2006}\)

\(Vì\dfrac{1}{2004.2005}>\dfrac{1}{2005.2006}\Rightarrow1-\dfrac{1}{2004.2005}< 1-\dfrac{1}{2005.2006}\Rightarrow\dfrac{2004.2005-1}{2004.2005}< \dfrac{2005.2006-1}{2004.2006}\)

\(=\frac{2006.2004+2006-1}{2004.2006+2005}=\frac{2006.2004+2005}{2004.2006+2005}=1\)

Việt làm đúng rồi nên mik ko Cần giải cho bạn nữa ,OK?

=2006×(2004+1)-1/2004×2006+2005

=2006×2004+2006×1-1/2004×2006+2005

=2006×2004+2005/2004×2006+2005

=1

= 1/1 - 1/5 + 1/5 -1/6 + 1/6 - 1/7 + 1/7 - 1/8 + 1/8-1/9 +...+ 1/2004 - 1/2005

= 1/1 - 1/2005

= 2004/2005

ai giúp mình vời

thế kết quả thứ 2 của mình là 2006/2005

\(a,\left(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right):\frac{1}{4}:\frac{1}{6}\)

\(=\left(\frac{1}{6}+\frac{1}{4}-\frac{1}{5}\right)\cdot\frac{1}{4}\cdot\frac{1}{6}\)

\(=\left(\frac{10}{60}+\frac{15}{60}-\frac{12}{60}\right)\cdot\frac{1}{24}\)

\(=\frac{13}{60}\cdot\frac{1}{24}\)

\(=\frac{13}{1440}\)

\(b,\frac{2006\cdot2005-1}{2004\cdot2006+2005}\)

 \(\frac{2006\cdot2005-1}{2004\cdot2006+2005}\)

\(=\frac{2006\cdot\left(2004+1\right)-1}{2004 \cdot2006+2005}\)

\(=\frac{2006\cdot2004+2006\cdot1-1}{2004\cdot2006+2005}\)

\(=\frac{2006\cdot2004+2005}{2004\cdot2006 +2005}=1\)

Mình nghĩ phần b, ko có cách 2 đâu bạn .

Bài làm:

Ta có: \(\frac{2004.2006-2003}{2005.2005-2004}\)

\(=\frac{\left(2005-1\right)\left(2005+1\right)-2003}{2005.2005-2004}\)

\(=\frac{2005.2005+2005-2005-1-2003}{2005.2005-2004}\)

\(=\frac{2005.2005-2004}{2005.2005-2004}\)

\(=1\)

\(\frac{2004.2006-2003}{2005.2005-2004}\)=\(\frac{2004.2005+2004-2003}{2005.2004+2005-2004}\)

                                            =\(\frac{2004.2005+1}{2005.2004+1}\)

                                            =1

Chúc bạn học tốt

Ta có : 

\(\frac{2004.2006-2003}{2005.2005-2004}=\frac{\left(2005-1\right).\left(2005+1\right)-2003}{2005.2005-2004}=\frac{2005.2005-2005+2005-1-2003}{2005.2005-2004}\)

 \(=\frac{2005.2005-2004}{2005.2005-2004}=1\)

\(\frac{\left(2005+1\right).125+1000}{\left(125+1\right).2005-888}\)

= \(\frac{2005.125+125+1000}{125.2005+2005-888}\)

= \(\frac{2005.125+1125}{125.2005+1117}\)

= \(\frac{250625+1125}{250625+1117}\)

= \(\frac{125875}{125871}\)

\(\dfrac{2004\cdot2007+6}{2005\cdot2005+2009}=\dfrac{2004\cdot2005+2004\cdot2+6}{2005\cdot2004+2005+2009}\\ =\dfrac{2004\cdot2005+4014}{2004\cdot2005+4014}=1\)