A = \(\dfrac{6\sqrt{x}+8}{3\sqrt{x}+2}=2+\dfrac{4}{3\sqrt{x}+2}\)
Có \(3\sqrt{x}+2>0< =>\dfrac{4}{3\sqrt{x}+2}>0\) <=> A > 2
Có: \(3\sqrt{x}+2\ge2< =>\dfrac{4}{3\sqrt{x}+2}\le2\) <=> A \(\le4\)
<=> 2 < A \(\le4\)
Mà A nguyên
<=> \(\left[{}\begin{matrix}A=3\\A=4\end{matrix}\right.\)
TH1: A = 3
<=> \(\dfrac{4}{3\sqrt{x}+2}=1\)
<=> \(3\sqrt{x}+2=4< =>x=\dfrac{4}{9}\)
TH2: A = 4
<=> \(\dfrac{4}{3\sqrt{x}+2}=2< =>3\sqrt{x}+2=2< =>x=0\)