Question

Tìm x nguyên để \(\dfrac{2\sqrt{x}+5}{\sqrt{x}-2}\) nguyên

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

Để \(\dfrac{2\sqrt{x}+5}{\sqrt{x}-2}\) là số nguyên thì \(2\sqrt{x}+5⋮\sqrt{x}-2\)

\(\Leftrightarrow2\sqrt{x}-4+9⋮\sqrt{x}-2\)

mà \(2\sqrt{x}-4⋮\sqrt{x}-2\)

nên \(9⋮\sqrt{x}-2\)

\(\Leftrightarrow\sqrt{x}-2\inƯ\left(9\right)\)

\(\Leftrightarrow\sqrt{x}-2\in\left\{1;-1;3;-3;9;-9\right\}\)

mà \(\sqrt{x}-2\ge-2\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{x}-2\in\left\{1;-1;3;9\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{3;1;5;11\right\}\)

hay \(x\in\left\{9;1;25;121\right\}\)(nhận)

Vậy: Để \(\dfrac{2\sqrt{x}+5}{\sqrt{x}-2}\) nguyên thì \(x\in\left\{9;1;25;121\right\}\)