\(\dfrac{9}{382};\dfrac{6}{257};\dfrac{15}{643}\)
\(BCNN\left(9,6,15\right)=90.\)
\(\Rightarrow\) Ta có:
\(\dfrac{9}{382}=\dfrac{9\times10}{382\times10}=\dfrac{90}{3820}.\)
\(\dfrac{6}{257}=\dfrac{6\times15}{257\times15}=\dfrac{90}{3855}.\)
\(\dfrac{15}{643}=\dfrac{15\times6}{643\times6}=\dfrac{90}{3858}.\)
Mà \(\dfrac{90}{3820}>\dfrac{90}{3855}>\dfrac{90}{3858}\). Do đó \(\dfrac{9}{382}>\dfrac{6}{257}>\dfrac{15}{643.}\)
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