\(\Leftrightarrow a=\dfrac{24-4b}{3}\)
để a,b là số tự nhiên khi \(\left(24-4b\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow\left[{}\begin{matrix}24-4b=1\\24-4b=-1\\24-4b=3\\24-4b=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}b=\dfrac{23}{4}\left(l\right)\\b=\dfrac{25}{4}\left(l\right)\\b=\dfrac{21}{4}\left(l\right)\\b=\dfrac{27}{4}\left(l\right)\end{matrix}\right.\)
vậy ko tồn taị a,b
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