$\displaystyle \text{ta có } \frac{1}{\sqrt{n} + \sqrt{n+2}} = \frac{\sqrt{n+2} - \sqrt{n}}{2} \quad (\forall n > 0)$

$\displaystyle \text{Đặt } B = \frac{1}{\sqrt{3} + \sqrt{5}} + \frac{1}{\sqrt{7} + \sqrt{9}} + \cdots + \frac{1}{\sqrt{143} + \sqrt{145}}$

$\displaystyle \Rightarrow A > B$

$\displaystyle \Rightarrow 2A > A + B = \left( \frac{1}{\sqrt{1} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{5}} \right) + \cdots + \left( \frac{1}{\sqrt{141} + \sqrt{143}} + \frac{1}{\sqrt{143} + \sqrt{145}} \right)$

$\displaystyle \Rightarrow 2A > \frac{\sqrt{3} - \sqrt{1}}{2} + \frac{\sqrt{5} - \sqrt{3}}{2} + \cdots + \frac{\sqrt{145} - \sqrt{143}}{2}$

$\displaystyle \Rightarrow 2A > \frac{\sqrt{145} - 1}{2} \Rightarrow A > \frac{\sqrt{145} - 1}{4}$

$\displaystyle \text{Vì } \sqrt{145} > \sqrt{144} = 12 \Rightarrow A > \frac{12 - 1}{4} = 2.75 > 2 \quad (1)$

$\displaystyle \text{Đặt } C = \frac{1}{\sqrt{3} + \sqrt{5}} + \frac{1}{\sqrt{7} + \sqrt{9}} + \cdots + \frac{1}{\sqrt{139} + \sqrt{141}}$

$\displaystyle \Rightarrow A - \frac{1}{\sqrt{1} + \sqrt{3}} < C$

$\displaystyle \Rightarrow 2\left(A - \frac{1}{\sqrt{1} + \sqrt{3}}\right) < \left(A - \frac{1}{\sqrt{1} + \sqrt{3}}\right) + C$

$\displaystyle \Rightarrow 2A - (\sqrt{3} - 1) < \frac{1}{\sqrt{3} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{7}} + \cdots + \frac{1}{\sqrt{141} + \sqrt{143}}$

$\displaystyle \Rightarrow 2A - \sqrt{3} + 1 < \frac{\sqrt{5} - \sqrt{3}}{2} + \frac{\sqrt{7} - \sqrt{5}}{2} + \cdots + \frac{\sqrt{143} - \sqrt{141}}{2}$

$\displaystyle \Rightarrow 2A - \sqrt{3} + 1 < \frac{\sqrt{143} - \sqrt{3}}{2}$

$\displaystyle \Rightarrow 2A < \frac{\sqrt{143} - \sqrt{3}}{2} + \sqrt{3} - 1 = \frac{\sqrt{143} + \sqrt{3} - 2}{2}$

$\displaystyle \Rightarrow A < \frac{\sqrt{143} + \sqrt{3} - 2}{4}$

$\displaystyle \text{Vì } \sqrt{143} < \sqrt{144} = 12 \text{ và } \sqrt{3} < 2 \Rightarrow A < \frac{12 + 2 - 2}{4} = 3 \quad (2)$

$\displaystyle \text{Từ } (1) \text{ và } (2) \Rightarrow 2 < A < 3$

$\displaystyle \Rightarrow A \notin \mathbb{Z} \quad \text{(đpcm)}.$