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\(\frac{\left(2005+1\right).125+1000}{\left(125+1\right).2005-888}\)
= \(\frac{2005.125+125+1000}{125.2005+2005-888}\)
= \(\frac{2005.125+1125}{125.2005+1117}\)
= \(\frac{250625+1125}{250625+1117}\)
= \(\frac{125875}{125871}\)
\(=15x\left(\dfrac{21}{43}+\dfrac{22}{43}\right)=15x1=15\)
A = 2006 x 125 + 1000 / 126 x 2005 - 888
A = (2005 + 1) x 125 + 1000 / (125 + 1) x 2005 - 888
A = 2005 x 125 + 125 + 1000 / 125 x 2005 + 2005 - 888
A = 2005 x 125 + 1125 / 125 x 2005 + 1117
A = 1125 / 1117
=13/12x14/13x15/14x16/15x...x2006/2005x2007/2006x2008/2007
=2008/12
=502/3
A = 1\(\dfrac{1}{12}\) \(\times\) 1\(\dfrac{1}{13}\) \(\times\) 1\(\dfrac{1}{14}\) \(\times\) 1\(\dfrac{1}{15}\) \(\times\) ... \(\times\) 1\(\dfrac{1}{2005}\) \(\times\) 1\(\dfrac{1}{2006}\) \(\times\) 1\(\dfrac{1}{2007}\)
A = ( 1 + \(\dfrac{1}{12}\)) \(\times\) ( 1 + \(\dfrac{1}{13}\)) \(\times\) ( 1 + \(\dfrac{1}{14}\)) \(\times\)...\(\times\) ( 1 + \(\dfrac{1}{2006}\))\(\times\)(1+\(\dfrac{1}{2007}\))
A = \(\dfrac{13}{12}\) \(\times\) \(\dfrac{14}{13}\) \(\times\) \(\dfrac{15}{14}\) \(\times\) ...\(\times\) \(\dfrac{2007}{2006}\) \(\times\) \(\dfrac{2008}{2007}\)
A = \(\dfrac{13\times14\times15\times...\times2007}{13\times14\times15\times...\times2007}\) \(\times\) \(\dfrac{2008}{12}\)
A = 1 \(\times\) \(\dfrac{502}{3}\)
A = \(\dfrac{502}{3}\)
\(1,\\ a,=\dfrac{7}{19}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)=\dfrac{7}{19}\times1=\dfrac{7}{19}\\ b,=\dfrac{2}{5}+\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=\dfrac{2}{5}+1=\dfrac{7}{5}\\ 2,\\ a,=15\times\left(\dfrac{2121}{4343}+\dfrac{222222}{434343}\right)\\ =15\times\left(\dfrac{2121:101}{4343:101}+\dfrac{222222:10101}{434343:10101}\right)\\ =15\times\left(\dfrac{21}{43}+\dfrac{22}{43}\right)=15\times\dfrac{43}{43}=15\times1=15\)
\(3,\)
Cạnh \(AC=\) chu vi ABC \(-AB-BC=\dfrac{4}{5}-\dfrac{1}{5}-\dfrac{1}{4}=\dfrac{3}{5}-\dfrac{1}{4}=\dfrac{7}{20}\left(m\right)\)
Vì \(\dfrac{7}{20}>\dfrac{5}{20}>\dfrac{4}{20}\Rightarrow\dfrac{7}{20}>\dfrac{1}{4}>\dfrac{1}{5}\) nên \(AC>BC>AB\)
\(a,=15\left(\dfrac{2121}{4343}+\dfrac{222222}{434343}\right)=15\left(\dfrac{21}{43}+\dfrac{22}{43}\right)=15\cdot1=15\)