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 : 

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

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

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

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

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

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

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

Ta có:

\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)

\(\Rightarrow\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2005.2006.2007}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+...+\frac{1}{2005.2006}-\frac{1}{2006.2007}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2006.2007}\right)=\frac{1}{2}\left(\frac{2005.2008}{2.2006.2007}\right)\)

Đặt \(A=1.2+2.3+...+n\left(n+1\right)\)

\(\Rightarrow3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+...+n\left(n+1\right)\left(n+2-\left(n-1\right)\right)\)

\(\Rightarrow3A=1.2.3-1.2.0+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)

\(\Rightarrow3A=n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

\(\Rightarrow1.2+2.3+...+2006.2007=\frac{2006.2007.2008}{2}\)

Vậy pt trở thành:

\(\frac{1}{2}\left(\frac{2005.2008}{2.2006.2007}\right)x=\frac{2006.2007.2008}{2}\)

\(\Leftrightarrow\frac{2005}{2.2006.2007}x=2006.2007\)

\(\Rightarrow x=\frac{2.\left(2006.2007\right)^2}{2005}\)

=2006×(2004+1)-1/2004×2006+2005

=2006×2004+2006×1-1/2004×2006+2005

=2006×2004+2005/2004×2006+2005

=1

dùng hàng đẳng thức bình phương tổng 2 số là auto ra, cái chính là tách khéo léo để tạo được thành hàng đẳng thức nhá !!!

a) \(498^2+996.502+502^2\)

\(=498^2+2.498.502+502^2\)

\(=\left(498+502\right)^2\)

\(=1000^2\)

\(=1000000\)

b) \(126^2-52.126+26^2\)

\(=126^2-2.26.126+26^2\)

\(=\left(126-26\right)^2\)

\(=100^2\)

\(=10000\)

c) \(1995^2-1994.1996\)

\(=1995^2-\left(1995-1\right)\left(1995+1\right)\)

\(=1995^2-\left(1995^2-1\right)\)

\(=1995^2-1995^2+1\)

\(=1\)

d) \(2005^2-2004.2006\)

\(=2005^2-\left(2005-1\right)\left(2005+1\right)\)

\(=2005^2-\left(2005^2-1\right)\)

\(=2005^2-2005^2+1\)

\(=1\)

4020022,999 nhe bn kik nha cam on nhieu

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