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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
Đáp án:
244x395-151
244+395x243
=243x395+395-151
243x395+244
=243x395+244
243x395+244
=1
Đáp án:
244x395-151
244+395x243
=243x395+395-151
243x395+244
=243x395+244
243x395+244
=1
tk cho mik nhé
\(A=\frac{2005}{2006}+\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2005}=1-\frac{1}{2006}+1-\frac{1}{2007}+1-\frac{1}{2008}+1+\frac{1}{2005}\)
\(=\left(1+1+1+1\right)+\left(\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}-\frac{1}{2008}\right)<4+\left(\frac{3}{6015}-3.\frac{1}{2006}\right)\)
\(=4+\left(\frac{3}{6015}-\frac{3}{2006}\right)<4+\left(\frac{3}{2006}-\frac{3}{2006}\right)=4+0=4\)
=>đpcm
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2004\cdot2005}+\frac{1}{2005\cdot2006}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2004}-\frac{1}{2005}+\frac{1}{2005}-\frac{1}{2006}\)
\(A=1-\frac{1}{2006}=\frac{2005}{2006}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2005.2006}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(\Rightarrow A=1-\frac{1}{2006}\)
\(\Rightarrow A=\frac{2005}{2006}\)
A=\(\frac{2006x\left(2004+1\right)-1}{2006x2004+2005}=\frac{2006x2004+2006-1}{2006x2004+2005}\)
= \(\frac{2006x2004+2005}{2006x2004+2005}=1\)
\(a,\frac{15}{18}=\frac{5}{6}\)
\(b,\frac{144}{351}=\frac{16}{39}\)
\(c,\frac{6}{30}=\frac{1}{5}\)
\(d,\frac{2005}{3000}=\frac{401}{600}\)