\(b)ĐK:\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(B=\dfrac{2\sqrt{x}-5}{\sqrt{x}-1}=\dfrac{2\left(\sqrt{x}-1\right)-3}{\sqrt{x}-1}\)
\(=2-\dfrac{3}{\sqrt{x}-1}\)
\(B\in Z\Leftrightarrow\dfrac{3}{\sqrt{x}-1}\in Z\Leftrightarrow\sqrt{x}-1\inƯ\left(3\right)=\left\{\pm1;3\right\}\) (vì \(\sqrt{x}-1\ge-1,\sqrt{x}-1\in Z\))
Xét các TH:
\(TH1:\sqrt{x}-1=-1\)
\(\Leftrightarrow\sqrt{x}=0\)
\(\Leftrightarrow x=0\left(tm\right)\)
\(TH2:\sqrt{x}-1=1\)
\(\Leftrightarrow\sqrt{x}=2=\sqrt{4}\)
\(\Leftrightarrow x=4\left(tm\right)\)
\(TH3:\sqrt{x}-1=3\)
\(\Leftrightarrow\sqrt{x}=4=\sqrt[]{16}\)
\(\Leftrightarrow x=16\left(tm\right)\)
Vậy, \(B\in Z\Leftrightarrow x\in\left\{1;4;16\right\}\)
