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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
\(A=\frac{2005}{2006}+\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2005}=1-\frac{1}{2006}+1-\frac{1}{2007}+1-\frac{1}{2008}+1+\frac{1}{2005}\)
\(=\left(1+1+1+1\right)+\left(\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}-\frac{1}{2008}\right)<4+\left(\frac{3}{6015}-3.\frac{1}{2006}\right)\)
\(=4+\left(\frac{3}{6015}-\frac{3}{2006}\right)<4+\left(\frac{3}{2006}-\frac{3}{2006}\right)=4+0=4\)
=>đpcm
\(a,\frac{15}{18}=\frac{5}{6}\)
\(b,\frac{144}{351}=\frac{16}{39}\)
\(c,\frac{6}{30}=\frac{1}{5}\)
\(d,\frac{2005}{3000}=\frac{401}{600}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2004\cdot2005}+\frac{1}{2005\cdot2006}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2004}-\frac{1}{2005}+\frac{1}{2005}-\frac{1}{2006}\)
\(A=1-\frac{1}{2006}=\frac{2005}{2006}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2005.2006}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(\Rightarrow A=1-\frac{1}{2006}\)
\(\Rightarrow A=\frac{2005}{2006}\)
1)2007/2006 > 1
2006/2008 < 1
2007/2006 > 1 > 2006/2008
2007/2006 >2006 /2008
2)1313/1414 = 1313: 101/1414 : 101 =13/14
13/14 = 13/14 => 1313/1414= 13/14
3) 97/96 - 1 = 1/96
96/95 - 1 = 1/95
1/96<1/95 => 97/96 < 96/95
4) 2007/2006 - 1 = 1/2006
2005/2004 - 1 = 1/2004
1/2006 < 1/2004 => 2007/2006 < 2005/2004
5) 2007/2006 - 1 = 1/2006
2008/2007 - 1 = 1/2007
1/2006 > 1/2007 => 2007/2006 > 2008/2007
1) so sánh với 1
2) rút gọn rồi so sánh
3,4,5 ) dùng phần bù ( phân số có phần bù lớn hơn thì phân số đó lớn hơn )
a)\(\frac{19}{20}+\frac{1}{20}=1\)
\(\frac{20}{21}+\frac{1}{21}=1\)
vi \(\frac{1}{20}>\frac{1}{21}\) nen \(\frac{19}{20}<\frac{20}{21}\)
b) \(\frac{89}{88}-\frac{1}{88}=1\)
\(\frac{90}{89}-\frac{1}{89}=1\)
vi \(\frac{1}{88}>\frac{1}{89}nen\frac{89}{88}>\frac{90}{89}\)
c)\(\frac{2005}{2003}-\frac{2}{2003}=1\)
\(\frac{2003}{2001}-\frac{2}{2001}=1\)
vi \(\frac{2}{2003}<\frac{2}{2001}nen\frac{2005}{2003}<\frac{2003}{2001}\)
A=\(\frac{2006x\left(2004+1\right)-1}{2006x2004+2005}=\frac{2006x2004+2006-1}{2006x2004+2005}\)
= \(\frac{2006x2004+2005}{2006x2004+2005}=1\)