Đáp án:
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Giải thích các bước giải:
Ta có:
$\dfrac{103}{105}=1-\dfrac{2}{105}$
$\dfrac{13}{15}=1-\dfrac{2}{15}$
$\dfrac{2}{105}<\dfrac{2}{15}$
$1-\dfrac{2}{105}>1-\dfrac{2}{15}$
$\to \dfrac{103}{105}>\dfrac{13}{15}$
Vậy $\dfrac{103}{105}>\dfrac{13}{15}$.
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Đặt $k=\dfrac{13}{27}\to 2k=\dfrac{26}{27}$
Đặt $n=\dfrac{7}{15}\to 2n=\dfrac{14}{15}$
$2k=\dfrac{26}{27}\to 2k=1-\dfrac{1}{27}$
$2n=\dfrac{14}{15}\to 2n=1-\dfrac{1}{15}$
$\dfrac{1}{27}<\dfrac{1}{15}$
$1-\dfrac{1}{27}>1-\dfrac{1}{15}$
$2k>2n\to k>n$
Vậy $\dfrac{13}{27}>\dfrac{7}{15}$.
