Ответ:

  1. При \(a\ge0\): \(\sqrt{16a}+\sqrt{100a}-\sqrt{81a}=4\sqrt a+10\sqrt a-9\sqrt a=5\sqrt a\).
  2. \(\sqrt{20}-\sqrt{125}+\sqrt{405}=2\sqrt5-5\sqrt5+9\sqrt5=6\sqrt5\).
  3. При \(b\ge0\): \(4\sqrt{27b}-5\sqrt{48b}+\frac14\sqrt{192b}=4\cdot3\sqrt{3b}-5\cdot4\sqrt{3b}+\frac14\cdot8\sqrt{3b}=12\sqrt{3b}-20\sqrt{3b}+2\sqrt{3b}=-6\sqrt{3b}\).