ta có : A=53x107+107-53/53x107+54=53x107+54/53x107+54=1

           B=134x269+269-133/134x269+135=134x269+136/134x269+135>1(vì 136>135)

vậy A<B

k cho mình nhé

\(\dfrac{54\times107-53}{53\times107+54}=\dfrac{53\times107+53-53}{53\times107+54}=\dfrac{53\times107}{53\times107+54}< 1\)

\(\dfrac{135\times269-133}{134\times269+135}=\dfrac{134\times269+269-133}{134\times269+135}=\dfrac{134\times269+136}{134\times269+135}>1\)

\(\Rightarrow\dfrac{54\times107-53}{53\times107+54}< \dfrac{135\times269-133}{134\times269+135}\)

a=5778-107+54

a=5671+54

a=5725

Bạn nên sửa đề thành: \(\frac{54\cdot107-53}{53\cdot107+54}\)

Thì \(=\frac{54\cdot\left(54+53\right)-53}{53\left(53+54\right)+54}=\frac{54^2+53\left(53+1\right)-53}{53^2+54\cdot\left(54-1\right)+54}=\frac{54^2+53^2}{53^2+54^2}=1\)

Chứ nguyên đề thì tính hợp lý sao?

có nhầm đề ko

TA có :

A = \(\dfrac{54.107-53}{53.107+54}\)= \(\dfrac{53.107+107-53}{53.107+54}\)=

\(\dfrac{53.107+54}{53.107+54}\)= 1 (1)

B = \(\dfrac{135.269-133}{134.269+135}\)=

\(\dfrac{134.269-269-133}{134.269-133}\)= \(\dfrac{134.269+136}{134.269+135}\)>1 (2)

Từ (1) và (2)

=> A > B

Ta có: \(A=\dfrac{54\cdot107-53}{53\cdot107+54}=\dfrac{53\cdot107+157-53}{53\cdot107+54}=\dfrac{53\cdot107+54}{53\cdot107+54}=1\) (1)

\(B=\dfrac{135\cdot269-133}{134\cdot269+135}=\dfrac{134\cdot269+269-133}{134\cdot269+135}=\dfrac{134\cdot269+136}{134\cdot269+135}>1\) (2)

Từ (1) và (2) suy ra: \(A< B\)

Vậy \(A< B\).