`(\sqrtx+1)/(\sqrtx-3)` là số nguyên
`<=> \sqrtx +1 vdots \sqrtx-3`
`<=> (\sqrtx-3)+4 vdots \sqrtx-3`
`<=> \sqrtx-3 \in {-4;4;-2;2;-1;1}`
`<=> \sqrtx \in {7;1;5;2;4}`
`<=> x \in {49;1;25;4;16}`
Để \(A=\dfrac{\sqrt{x}+1}{\sqrt{3}-3}\) nguyên thì \(\sqrt{x}-1⋮\sqrt{3}-3\) \(\Leftrightarrow\) \(\sqrt{x}-1\in B\left(\sqrt{3}-3\right)\)
\(\Rightarrow\sqrt{x}-1=\left(\sqrt{3}-3\right)k\left(k\in Z\right)\)
\(\Leftrightarrow\sqrt{x}=\left(\sqrt{3}-3\right)k+1\)
\(\Rightarrow x=\left(\left(\sqrt{3}-3\right)k+1\right)^2\)
Vậy x sẽ có dạng \(x=\left(\left(\sqrt{3}-3\right)k+1\right)^2\) thì A\(⋮\sqrt{3}-3\)
Để biểu thức nhận giá trị nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{1;-1;2;-2;4;-4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{4;2;5;1;7;-1\right\}\)
\(\Leftrightarrow x\in\left\{16;4;25;1;49\right\}\)
