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Ta có : \(\frac{124124}{125125}=\frac{124124:1001}{125125:1001}=\frac{124}{125}\)
\(\frac{123}{124}=1-\frac{1}{124}\); \(\frac{124}{125}=1-\frac{1}{125}\)
Vì 1/124 < 1/125 => 123/124 > 123/125 => 123/124 > 124124/125125
ta có: \(\frac{124124}{125125}=\frac{124}{125}\)
Lại có: \(1-\frac{123}{124}=\frac{1}{124};1-\frac{124}{125}=\frac{1}{125}\)
\(\Rightarrow\frac{1}{124}>\frac{1}{125}\Rightarrow1-\frac{123}{124}>1-\frac{124}{125}\)
\(\Rightarrow\frac{123}{124}< \frac{124}{125}\Rightarrow\frac{123}{124}< \frac{124124}{125125}\)
Lời giải:
$\frac{1234}{1334}=1-\frac{100}{1334}> 1-\frac{100}{1330}=1-\frac{10}{133}=\frac{123}{133}$
A=\(\frac{54.107-53}{53.107+54}\)
= \(\frac{54.\left(106+1\right)-53}{53.107+53+1}\)
= \(\frac{54.106+54-53}{53.\left(107+1\right)+1}\)
= \(\frac{54.106+1}{53.108+1}\)
mà \(54.106=53.108\)
=> \(\frac{54.106+1}{53.108+1}=1\)
B = \(\frac{135.269-133}{134.269+135}\)
= \(\frac{135.\left(268+1\right)-133}{134.269+134+1}\)
= \(\frac{135.268+135-133}{134.\left(269+1\right)+1}\)
= \(\frac{135.268+2}{134.270+1}\)
Mà \(135.268=134.270\)
=> \(\frac{135.268+2}{134.270+1}=\frac{135.268+1+1}{134.270+1}\)
\(=\frac{135.268+1}{134.270+1}+\frac{1}{134.270+1}\)
\(=1+\frac{1}{134.270+1}>1\)
=> B > 1
=> B > A
\(A=\frac{54\cdot107-53}{53\cdot107+54}=\frac{\left(53+1\right)107-53}{53\cdot107+54}=\frac{53\cdot107+107-53}{53\cdot107+54}=\frac{53\cdot107+54}{53\cdot107+54}=1\)
\(B=\frac{135\cdot268-133}{134\cdot269+135}=\frac{\left(134+1\right)\cdot268-133}{134\cdot269+135}=\frac{134\cdot268+268-133}{34\cdot269+135}=\frac{134\cdot268+135}{134\cdot269+135}=1\)
Vì 1=1 nên A=B
\(A=\frac{\left(53+1\right).107-53}{53.107+54}=\frac{53.107+107-53}{53.107+54}=\frac{53.107+54}{53.107+54}=1.\)
\(B=\frac{\left(134+1\right).269-133}{134.269+135}=\frac{134.269+269-133}{134.269+135}=\frac{134.269+136}{134.269+135}>1\)
\(\Rightarrow A< B\)
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