a) Ta có: \(P = \frac{{\sqrt x }}{{\sqrt x {\rm{\;}} - 2}} + \frac{{\sqrt x }}{{\sqrt x {\rm{\;}} + 2}} - \frac{{x - 2\sqrt x }}{{x - 4}}\) \(P = \frac{{\sqrt x }}{{\sqrt x {\rm{\;}} - 2}} + \frac{{\sqrt x }}{{\sqrt x {\rm{\;}} + 2}} - \frac{{\sqrt x \left( {\sqrt x {\rm{\;}} - 2} \right)}}{{\left( {\sqrt x {\rm{\;}} - 2} \right)\left( {\sqrt x {\rm{\;}} + 2} \right)}}\) \(P = \frac{{\sqrt x }}{{\sqrt x {\rm{\;}} - 2}} + \frac{{\sqrt x }}{{\sqrt x {\rm{\;}} + 2}} - \frac{{\sqrt x }}{{\sqrt x {\rm{\;}} + 2}}\) \(P = \frac{{\sqrt x }}{{\sqrt x {\rm{\;}} - 2}}\). b) Thay \(x = 16\) vào P, ta được: \(P = \frac{{\sqrt {16} }}{{\sqrt {16}  - 2}} = \frac{4}{{4 - 2}} = \frac{4}{2} = 2\). Vậy với \(x = 16\) thì \(P = 2\). c) Ta có: \(M = P:Q = \frac{{\sqrt x }}{{\sqrt x {\rm{\;}} - 2}}:\frac{{\sqrt x {\rm{\;}} + 2}}{{\sqrt x {\rm{\;}} - 2}}{\mkern 1mu} \) \( = \frac{{\sqrt x }}{{\sqrt x {\rm{\;}} - 2}}.\frac{{\sqrt x {\rm{\;}} - 2}}{{\sqrt x {\rm{\;}} + 2}} = \frac{{\sqrt x }}{{\sqrt x {\rm{\;}} + 2}}\) Vì \({M^2} < \frac{1}{4}\) nên \({\left( {\frac{{\sqrt x }}{{\sqrt x {\rm{\;}} + 2}}} \right)^2} < \frac{1}{4}\). Suy ra \(\left| {\frac{{\sqrt x }}{{\sqrt x {\rm{\;}} + 2}}} \right| < \frac{1}{2}\) Vì \(\sqrt x  > 0\) nên \(\frac{{\sqrt x }}{{\sqrt x {\rm{\;}} + 2}} > 0\) Do đó \(\frac{{\sqrt x }}{{\sqrt x {\rm{\;}} + 2}} < \frac{1}{2}\) \(2\sqrt x {\rm{\;}} < \sqrt x {\rm{\;}} + 2\) \(\sqrt x {\rm{\;}} < 2\) \(x < 4\) Kết hợp điều kiện \(x \ge 0;x \ne 4\) ta được \(0 \le x < 4\). Vậy để \({M^2} < \frac{1}{4}\) thì \(0 \le x < 4\).