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\(A=\frac{54\cdot107-53}{53\cdot107+54}=\frac{\left(53+1\right)107-53}{53\cdot107+54}=\frac{53\cdot107+107-53}{53\cdot107+54}=\frac{53\cdot107+54}{53\cdot107+54}=1\)
\(B=\frac{135\cdot268-133}{134\cdot269+135}=\frac{\left(134+1\right)\cdot268-133}{134\cdot269+135}=\frac{134\cdot268+268-133}{34\cdot269+135}=\frac{134\cdot268+135}{134\cdot269+135}=1\)
Vì 1=1 nên A=B
\(A=\frac{54.107-53}{53.107+54}=\frac{\left(53+1\right).107-53}{53.107+54}\)
\(=\frac{53.107+107-53}{53.107+54}=\frac{53.107+54}{53.107+54}=1\)
\(B=\frac{135.269-133}{134.269+135}=\frac{\left(134+1\right).269-133}{134.269+135}\)
\(=\frac{134.269+269-133}{134.269+135}=\frac{134.269+136}{134.269+135}\)
\(=\frac{134.269+135}{134.269+135}+\frac{1}{134.269+135}=1+\frac{1}{134.269+135}\)
Vì \(1+\frac{1}{134.269+135}>1\Rightarrow A< B\)
Vậy \(A< B\)
A= (54.107-53)/(53.107+54)
= (53+1).107-53 / 53.107+54
=53.107+107-53 / 53.107+54
=53.107+54 / 54.107 + 54
=1
B= 135.269-133 / 134.269+135
= (134+1).269-133 / 134.269+135
= 134.269+269-133 / 134.269+135
=134.269+136 / 134.269+135
=134.269+135/ 134.269+135 + 1/134.269+135
=1 + 1/134.269+135 >1=A
\(A=\frac{54.107-53}{53.107+54}=1\)
\(B=\frac{135.269-133}{134.269+135}>1\)
\(A=\frac{54.107-53}{53.107+54}<\frac{135.269-133}{134.269+135}\)
\(M=\dfrac{54.107-53}{53.107-54}=\dfrac{53\left(107-1\right)+107}{53\left(107-1\right)-1}=\dfrac{53.106-1+108}{53.106-1}=1+\dfrac{108}{53.106-1}\)
\(N=\dfrac{135.269-133}{134.269-135}=\dfrac{134\left(269-1\right)-1+270}{134\left(269-1\right)-1}=1+\dfrac{270}{134.268-1}\)
\(M-N=\dfrac{108}{53.106-1}-\dfrac{270}{134.268-1}\)
\(M-N=\dfrac{2}{2.53^2-1}-\dfrac{5}{8.67^2-1}>\dfrac{5}{10.53^2-1}-\dfrac{5}{8.67^2-1}>0\)
M>N