a) \(P=\dfrac{2\sqrt{x}+12}{\sqrt{x}+2}=\dfrac{2\sqrt{x}+4+8}{\sqrt{x}+2}=\dfrac{2\left(\sqrt{x}+2\right)}{\sqrt{x}+2}+\dfrac{8}{\sqrt{x}+2}=2+\dfrac{8}{\sqrt{x}+2}\)
Với \(x\ge0\)
P nguyên khi:
\(\dfrac{8}{\sqrt{x}+2}\) nguyên
\(\Rightarrow8\) ⋮ \(\sqrt{x}+2\)
\(\Rightarrow\sqrt{x}+2\inƯ\left(8\right)=\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
Mà: \(\sqrt{x}+2\ge2\)
\(\Rightarrow\sqrt{x}+2\in\left\{2;4;8\right\}\)
\(\Rightarrow x\in\left\{0;4;36\right\}\)
b) \(P=\dfrac{4\sqrt{x}-7}{2\sqrt{x}-1}=\dfrac{4\sqrt{x}-2-5}{2\sqrt{x}-1}=\dfrac{2\left(2\sqrt{x}-1\right)}{2\sqrt{x}-1}-\dfrac{5}{2\sqrt{x}-1}=2-\dfrac{5}{2\sqrt{x}-1}\)
Với \(x\ge0\)
P nguyên khi:
\(\dfrac{5}{2\sqrt{x}-1}\) nguyên
\(\Rightarrow5\) ⋮ \(2\sqrt{x}-1\)
\(\Rightarrow2\sqrt{x}-1\inƯ\left(5\right)=\left\{5;-5;1;-1\right\}\)
Mà: \(2\sqrt{x}-1\ge-1\)
\(\Rightarrow2\sqrt{x}-1\in\left\{-1;1;5\right\}\)
\(\Rightarrow x\in\left\{0;1;9\right\}\)
