1: Thay x=9 vào Q, ta được:

\(Q = \frac{3 + 3}{3 + 1} = \frac{6}{4} = \frac{3}{2}\)

2: Sửa đề: \(P = \frac{1}{\sqrt{x} + 2} - \frac{\sqrt{x}}{1 - \sqrt{x}} - \frac{3 \sqrt{x}}{\left(\right. \sqrt{x} + 2 \left.\right) \left(\right. \sqrt{x} - 1 \left.\right)}\)

=>\(P = \frac{1}{\sqrt{x} + 2} + \frac{\sqrt{x}}{\sqrt{x} - 1} - \frac{3 \sqrt{x}}{\left(\right. \sqrt{x} + 2 \left.\right) \left(\right. \sqrt{x} - 1 \left.\right)}\)

\(= \frac{\sqrt{x} - 1 + x + 2 \sqrt{x} - 3 \sqrt{x}}{\left(\right. \sqrt{x} - 1 \left.\right) \left(\right. \sqrt{x} + 2 \left.\right)} = \frac{x - 1}{\left(\right. \sqrt{x} - 1 \left.\right) \left(\right. \sqrt{x} + 2 \left.\right)}\)

\(= \frac{\left(\right. \sqrt{x} - 1 \left.\right) \left(\right. \sqrt{x} + 1 \left.\right)}{\left(\right. \sqrt{x} - 1 \left.\right) \left(\right. \sqrt{x} + 2 \left.\right)} = \frac{\sqrt{x} + 1}{\sqrt{x} + 2}\)

3: \(M = P \cdot Q = \frac{\sqrt{x} + 3}{\sqrt{x} + 1} \cdot \frac{\sqrt{x} + 1}{\sqrt{x} + 2} = \frac{\sqrt{x} + 3}{\sqrt{x} + 2} = 1 + \frac{1}{\sqrt{x} + 2}\)

\(\frac{1}{\sqrt{x} + 2} < = \frac{1}{2} \forall x\) thỏa mãn ĐKXĐ

=>\(M = \frac{1}{\sqrt{x} + 2} + 1 < = \frac{3}{2} \forall x\) thỏa mãn ĐKXĐ