Cho biểu thức B = (a^2 - 3a căn bậc hai a + 2) / (a - 3 căn bậc hai x)

ĐKXĐ: a > 0, a ≠ 9.

B = \(\frac{{{a^2} - 3a\sqrt a + 2}}{{a - 3\sqrt a }} = \frac{{a\left( {a - 3\sqrt a } \right) + 2}}{{a - 3\sqrt a }} = a + \frac{2}{{a - 3\sqrt a }}\)

Để B nhận giá trị nguyên thì \(\frac{2}{{a - 3\sqrt a }} \in \mathbb{Z}\)

Suy ra: \(2 \vdots \left( {a - 3\sqrt a } \right)\)

⇒\(a - 3\sqrt a \in \left\{ { - 2; - 1;1;2} \right\}\)

⇔\(\left[ \begin{array}{l}a - 3\sqrt a + 2 = 0\\a - 3\sqrt a + 1 = 0\\a - 3\sqrt a - 1 = 0\\a - 3\sqrt a - 2 = 0\end{array} \right.\)

⇔\(\left[ \begin{array}{l}\left( {\sqrt a - 1} \right)\left( {\sqrt a - 2} \right) = 0\\\sqrt a = \frac{{3 \pm \sqrt 5 }}{2}\left( L \right)\\\sqrt a = \frac{{3 \pm \sqrt {13} }}{2}\left( L \right)\\\sqrt a = \frac{{3 \pm \sqrt {17} }}{2}\left( L \right)\end{array} \right.\)

⇔\(\left( {\sqrt a - 1} \right)\left( {\sqrt a - 2} \right) = 0\)

⇔\(\left[ \begin{array}{l}\sqrt a = 1\\\sqrt a = 2\end{array} \right.\)

⇔\(\left[ \begin{array}{l}a = 1\\a = 4\end{array} \right.\)

Vậy a = 1 hoặc a = 4.