So sánh M và \(\sqrt M \) biết M = \(\frac{{2 - 5\sqrt a }}{{\sqrt a + 3}}\) với a > 0.

\(\frac{{2 - 5\sqrt a }}{{\sqrt a + 3}} = \frac{{ - 5\left( {\sqrt a + 3} \right) + 15 + 2}}{{\sqrt a + 3}} = \frac{{ - 5\left( {\sqrt a + 3} \right) + 17}}{{\sqrt a + 3}} = - 5 + \frac{{17}}{{\sqrt a + 3}}\)

Vì \(\sqrt a \ge 0 \Rightarrow \sqrt a + 3 \ge 3 \Rightarrow \frac{{17}}{{\sqrt a + 3}} \le \frac{{17}}{3} \Rightarrow - 5 + \frac{{17}}{{\sqrt a + 3}} \le - 5 + \frac{{17}}{3} = \frac{2}{3}\)

Hay \(M < \frac{2}{3}\)

Suy ra: \(\sqrt M < M\).