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Ta có: \(\frac{2004\cdot2007+6}{2005\cdot2005+2009}=\frac{\left(2005-1\right)\cdot2007+6}{2005\cdot2005+2009}=\frac{2005\cdot2007-1\cdot2007+6}{2005\cdot2005+2009}=\frac{2005\cdot2007-2007+6}{2005\cdot2005+2009}\)
\(=\frac{\text{2005 x (2005 + 2) - 2007 + 6}}{\text{2005 x 2005 + 2009}}=\frac{\text{2005 x 2005 + 2005 x 2 - 2007 + 6}}{\text{2005 x 2005 + 2009}}=\frac{\text{2005 x 2005 + 4010 - 2007 + 6}}{\text{2005 x 2005 + 2009}}=\text{ }\frac{\text{2005 x 2005 + 2009}}{\text{2005 x 2005 + 2009}}=1\)
\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>\frac{2001}{2001}+\frac{2002}{2002}+\frac{2003}{2003}+\frac{2004}{2004}+\frac{2005}{2005}+\frac{2006}{2006}+\frac{2007}{2007}+\frac{2008}{2008}\)
\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>1+1+1+1+1+1+1+1\)\(A=\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>8\)
\(A>8\)
a. kết quả = 401/402
b. Ta có: 1-2004/2009=5/2009 , 1--2005/2010=5/2010 . Vì 5/2009 > 5/2010 nên 2004/2009 < 2005/2010.
Đấy phần b. mk ko quy đồng nha!
Nhớ Tích cho mk đấy
\(=\dfrac{\left(2005-1\right)\left(2005+2\right)+6}{2005^2+2005+4}\)
\(=\dfrac{2005^2+2005-2+6}{2005^2+2005+4}=1\)