1: Thay x=9 vào A, ta được:

\(A = \frac{3 \cdot 3}{3 + 2} = \frac{9}{5}\)

2: \(B = \frac{x + 4}{x - 4} - \frac{2}{\sqrt{x} - 2}\)

\(= \frac{x + 4}{\left(\right. \sqrt{x} - 2 \left.\right) \left(\right. \sqrt{x} + 2 \left.\right)} - \frac{2}{\sqrt{x} - 2}\)

\(= \frac{x + 4 - 2 \left(\right. \sqrt{x} + 2 \left.\right)}{\left(\right. \sqrt{x} - 2 \left.\right) \left(\right. \sqrt{x} + 2 \left.\right)} = \frac{x - 2 \sqrt{x}}{\left(\right. \sqrt{x} - 2 \left.\right) \left(\right. \sqrt{x} + 2 \left.\right)}\)

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

3: \(A - B < \frac{3}{2}\)

=>\(\frac{3 \sqrt{x}}{\sqrt{x} + 2} - \frac{\sqrt{x}}{\sqrt{x} + 2} < \frac{3}{2}\)

=>\(\frac{2 \sqrt{x}}{\sqrt{x} + 2} - \frac{3}{2} < 0\)

=>\(\frac{4 \sqrt{x} - 3 \left(\right. \sqrt{x} + 2 \left.\right)}{2 \left(\right. \sqrt{x} + 2 \left.\right)} < 0\)

=>\(\frac{\sqrt[]{x} - 6}{2 \left(\right. \sqrt{x} + 2 \left.\right)} < 0\)

=>\(\sqrt{x} - 6 < 0\)

=>\(\sqrt{x} < 6\)

=>0<=x<36

mà x là số nguyên dương lớn nhất thỏa mãn

nên x=35