\(A\sqrt{x}+A=9\Leftrightarrow\sqrt{x}=\dfrac{9-A}{A}\)
Ta có \(\sqrt{x}\ge0\Rightarrow\dfrac{9-A}{A}\ge0\)
TH1 : \(\left\{{}\begin{matrix}9-A>0\\9-A\ne0\\A>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}A< 9\\A\ne9\\A>0\end{matrix}\right.\Leftrightarrow0< A< 9\)
TH2 : \(\left\{{}\begin{matrix}9-A< 0\\9-A\ne0\\A< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}A>9\\A\ne9\\A< 0\end{matrix}\right.\left(voli\right)\)
=> \(x\in\left\{1;2;3;4;5;6;7;8\right\}\)thì A nguyên
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