=2006×(2004+1)-1/2004×2006+2005

=2006×2004+2006×1-1/2004×2006+2005

=2006×2004+2005/2004×2006+2005

=1

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?

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

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

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

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

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

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

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

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

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

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

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

\(\frac{2004.2005+2006.6-6}{2005.197+4.2005}\)= \(\frac{2004.2005+\left(2006-1\right).6}{2005.\left(197+4\right)}\)= \(\frac{2004.2005+2005.6}{2005.201}\)= \(\frac{\left(2004+6\right).2005}{2005.201}\)

= \(\frac{2010}{201}\)= \(10\)

\(P=\frac{\left(2003^2\cdot2013+31\cdot2004-1\right)\left(2003\cdot2008+4\right)}{2004\cdot2005\cdot2006\cdot2007\cdot2008}\)

Đặt a=2004 ta có

\(P=\frac{\left[\left(x-1\right)^2\cdot\left(a+9\right)+31\cdot a-1\right]\left[\left(a-1\right)\left(a+4\right)+4\right]}{a\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)}\)

\(=\frac{\left[\left(a^2-2a+1\right)\left(a+9\right)+31a-1\right]\left[\left(a^2+3a-4\right)+4\right]}{a\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)}\)

\(=\frac{\left(a^3+9a^2-2a^2-18a+a+9+31a-1\right)\left(a^2+3a\right)}{a\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)}\)

\(=\frac{\left(a^3+7a^2+14a+8\right)\left(a^2+3a\right)}{a\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)}\)

\(=\frac{a\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)}{a\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)}=1\)

Vậy \(P=1\)

Ui ko khó đâu chỉ lắm số thôi bạn ạ ~~~

Ta xét tử số: (2003^2.2013+31.2004-1)(2003.2008+4)

=[2003^2(2003+10)+(2003+1).31-1][2003(2003+5)+4]

=[2003^3+10.2003^2+31.2003+30][2003^2+5.2003+4]

Đặt 2003=a cho đỡ phức tạp

=(a^3+10a^2+31a+30)(a^2+5a+4)

Đến đây bạn phân tích đa thức thành nhân tử thôi

=(a+5)(a+2)(a+3)(a+1)(a+4)

Xét mẫu số khi đặt 2003=a

=> MS=(a+1)(a+2)(a+3)(a+4)(a+5)

=> P=1

Vậy P=1.

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

Đặt \(A=\dfrac{2003.2004-1}{2003.2004}\) và \(B=\dfrac{2004.2005-1}{2004.2005}\)

Ta có : \(A=\dfrac{2003.2004-1}{2003.2004}=\dfrac{2003.2004}{2003.2004}-\dfrac{1}{2003.2004}\)

\(=1-\dfrac{1}{2003.2004}\)

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

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

Vì \(\dfrac{1}{2003.2004}>\dfrac{1}{2004.2005}\Rightarrow1-\dfrac{1}{2003.2004}< 1-\dfrac{1}{2004.2005}\)

Nên \(A< B\)

Vậy \(\dfrac{2003.2004-1}{2003.2004}< \dfrac{2004.2005-1}{2004.2005}\)

~ Học tốt ~