A=54.107-53 phần 53.107+54
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\(A=\dfrac{54.107-53}{53.107+54}=\dfrac{53.107+107-53}{53.107+54}=\dfrac{53.107+54}{53.107+54}=1\\ B=\dfrac{135.169-133}{134.269+135}=\dfrac{135.169-133}{134.269+134+1}=\dfrac{135.169-133}{134.270+1}=\dfrac{135.169-133}{268.135+1}< 1\\ Vậy:B< A\left(A=1\right).hay.A>B\)
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=0,01033545893