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ta có: \(\frac{73}{75}>\frac{73}{79}>\frac{77}{79}\Rightarrow\frac{73}{75}>\frac{77}{79}\)
ta có: \(\frac{53}{100}< \frac{47}{100}\)
ta có: \(\frac{48}{47}>1;\frac{84}{85}< 1\Rightarrow\frac{48}{47}>\frac{84}{85}\)
\(\dfrac{47}{95}\) và \(\dfrac{35}{69}\)
\(\dfrac{47}{95}< \dfrac{1}{2}\) và \(\dfrac{35}{69}>\dfrac{1}{2}\)
Vậy \(\dfrac{47}{95}< \dfrac{35}{69}\)
\(\dfrac{53}{103}\) và \(\dfrac{71}{145}\)
\(\dfrac{53}{103}>\dfrac{1}{2}\) và \(\dfrac{71}{145}< \dfrac{1}{2}\)
Vậy \(\dfrac{53}{103}>\dfrac{71}{145}\)
\(\dfrac{2009}{2010}\) và \(\dfrac{2005}{2006}\)
\(1-\dfrac{2009}{2010}=\dfrac{1}{2010}\) và \(1-\dfrac{2005}{2006}=\dfrac{1}{2006}\)
Vậy \(\dfrac{2009}{2010}>\dfrac{2005}{2006}\)
\(\dfrac{783}{901}\) và \(\dfrac{738}{915}\)
\(\dfrac{738}{915}< \dfrac{783}{915}< \dfrac{783}{901}\)
Vậy \(\dfrac{783}{901}>\dfrac{738}{915}\)
`a)75<77`
`=>1/75>1/77`
`=>37/75>37/77>35/77`
`b)7779>7569`
`=>1/7779<1/7569`
`=>3734/7569>3734/7779>2099/7779`
\(\dfrac{27}{86}\) < \(\dfrac{27}{81}\) = \(\dfrac{1}{3}\) = \(\dfrac{9}{27}\) < \(\dfrac{25}{27}\)
Vậy \(\dfrac{27}{86}\) < \(\dfrac{25}{27}\)
\(\dfrac{51}{53}+\dfrac{55}{57}+\dfrac{61}{63}+\dfrac{69}{71}+\dfrac{79}{81}+\dfrac{91}{93}\)
\(=\left(\dfrac{52}{53}-\dfrac{1}{53}\right)+\left(\dfrac{56}{57}-\dfrac{1}{57}\right)+\left(\dfrac{62}{63}-\dfrac{1}{63}\right)+\left(\dfrac{70}{71}-\dfrac{1}{71}\right)+\left(\dfrac{80}{81}-\dfrac{1}{81}\right)+\left(\dfrac{92}{93}-\dfrac{1}{93}\right)\)
\(=\left(1-\dfrac{1}{53}-\dfrac{1}{53}\right)+\left(1-\dfrac{1}{57}-\dfrac{1}{57}\right)+\left(1-\dfrac{1}{63}-\dfrac{1}{63}\right)+\left(1-\dfrac{1}{71}-\dfrac{1}{71}\right)+\left(1-\dfrac{1}{81}-\dfrac{1}{81}\right)+\left(1-\dfrac{1}{93}-\dfrac{1}{93}\right)\)
\(=\left(1-0\right)+\left(1-0\right)+\left(1-0\right)+\left(1-0\right)+\left(1-0\right)+\left(1-0\right)\)
\(=1+1+1+1+1+1\)
\(=6\)
\(a,\dfrac{11}{49}< \dfrac{11}{46};\dfrac{11}{46}< \dfrac{13}{46}\\ Nên:\dfrac{11}{49}< \dfrac{13}{46}\\ b,\dfrac{62}{85}< \dfrac{62}{80};\dfrac{62}{80}< \dfrac{73}{80}\\ Nên:\dfrac{62}{85}< \dfrac{73}{80}\\ c,\dfrac{n}{n+3}< \dfrac{n}{n+2};\dfrac{n}{n+2}< \dfrac{n+1}{n+2}\\ Nên:\dfrac{n}{n+3}< \dfrac{n+1}{n+2}\)
1,
Ta có:
\(\dfrac{73}{75}=1-\dfrac{2}{75}\)
\(\dfrac{77}{79}=1-\dfrac{2}{79}\)
So sánh phân số \(\dfrac{2}{75}\) và \(\dfrac{2}{79}\)
Vì \(75< 79\) nên \(\dfrac{1}{75}>\dfrac{1}{79}\)
Vậy \(1-\dfrac{2}{75}< 1-\dfrac{2}{79}\)
Hay \(\dfrac{73}{75}< \dfrac{77}{79}\)
2,
Vì \(\dfrac{53}{100}>\dfrac{47}{100}>\dfrac{47}{106}\) nên \(\dfrac{53}{100}>\dfrac{47}{106}\)
3,
Ta có:
\(\dfrac{81}{79}=1+\dfrac{2}{79}\)
\(\dfrac{65}{63}=1+\dfrac{2}{63}\)
So sánh phân số \(\dfrac{2}{79}\) và \(\dfrac{2}{63}\)
Vì \(79>63\) nên \(\dfrac{81}{79}< \dfrac{65}{63}\)
Hay \(\Rightarrow1+\dfrac{2}{79}< 1+\dfrac{2}{63}\)
Vậy \(\dfrac{81}{79}< \dfrac{65}{63}\)
4,
\(\dfrac{48}{47}>1>\dfrac{84}{85}\)
Vậy \(\dfrac{48}{47}>\dfrac{84}{85}\)
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