Bài 1:
$Q = \frac{2 \sqrt {x} - 9}{x - 5\sqrt {x} + 6} - \frac{\sqrt {x} + 3}{\sqrt {x} - 2} - \frac{2\sqrt {x} + 1}{3 - \sqrt {x}}$
a.
+ ĐK: $\left\{ \begin{array}x x ≥ 0 \\ x ≠ 4 \\ x ≠ 9 \\ \end{array} \right.$
+ Ta có: $Q = \frac{2 \sqrt {x} - 9}{(\sqrt {x} - 2)(\sqrt {x} - 3)} - \frac{\sqrt {x} + 3}{\sqrt {x} - 2} - \frac{2\sqrt {x} + 1}{\sqrt {x} - 3}$
$= \frac{2\sqrt {x} - 9 - (x - 9) + (2\sqrt {x} + 1)(\sqrt {x} - 2)}{(\sqrt {x} - 2)(\sqrt {x} - 3)}$
$= \frac{2\sqrt {x} - 9 - x + 9 + 2x - 3\sqrt {x} - 2}{(\sqrt {x} - 2)(\sqrt {x} - 3)}$
$= \frac{x - \sqrt {x} - 2}{(\sqrt {x} - 3)(\sqrt {x} - 2)}$
$= \frac{(\sqrt {x} - 2)(\sqrt {x} + 1)}{(\sqrt {x} - 3)(\sqrt {x} - 2)}$
$= \frac{\sqrt {x} + 1}{\sqrt {x} - 3}$
b.
+ Ta có: $Q < 1 ⇔ \frac{\sqrt {x} + 1}{\sqrt {x} - 3} < 1$
$⇔ \frac{\sqrt {x} + 1}{\sqrt {x} - 3} - 1 < 0$
$⇔ \frac{\sqrt {x} + 1 - \sqrt {x} + 3}{\sqrt {x} - 3} < 0$
$⇔ \frac{4}{\sqrt {x} - 3} < 0$
$⇔ \sqrt {x} - 3 < 0$
$⇔ \sqrt {x} < 3$
$⇔ x < 9$
+ So với đk $x ≥ 0$ $⇒ 0 ≤ x < 9$.
c.
+ Ta có: $Q = \frac{\sqrt {x} + 1}{\sqrt {x} - 3} = \frac{\sqrt {x} - 3 + 4}{\sqrt {x} - 3} = 1+ \frac{4}{\sqrt {x} - 3}$
+ $Q$ nguyên $⇔ 4 \vdots \sqrt {x} - 3$
$⇔$ \(\left[ \begin{array}{l} \sqrt {x} - 3 = 1 \\ \sqrt {x} - 3 = - 1 \\ \sqrt {x} - 3 = 2 \\ \sqrt {x} - 3 = -2 \\ \sqrt {x} - 3 = 4 \end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l} x = 16 \\ x = 4 \\ x = 25 \\ x = 1 \\ x = 49\end{array} \right.\)
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