Ответ:

Используем свойства: \(\sqrt{ab}=\sqrt a\cdot\sqrt b\), приведение подобных радикалов и формулу \((a-b)(a+b)=a^2-b^2\).

  1. \(3\sqrt{72}-2\sqrt{200}+5\sqrt{32}-\sqrt{400}=3\cdot6\sqrt2-2\cdot10\sqrt2+5\cdot4\sqrt2-20\)
    \(=18\sqrt2-20\sqrt2+20\sqrt2-20=18\sqrt2-20\).
  2. \(4\sqrt{48}-5\sqrt{27}-\sqrt{300}+10\sqrt{108}=4\cdot4\sqrt3-5\cdot3\sqrt3-10\sqrt3+10\cdot6\sqrt3\)
    \(=16\sqrt3-15\sqrt3-10\sqrt3+60\sqrt3=51\sqrt3\).
  3. \(\sqrt5\cdot\sqrt{20}=\sqrt{100}=10\).
  4. \(\sqrt5\cdot\sqrt{45}=\sqrt{225}=15\).
  5. \(2\sqrt6\cdot3\sqrt6=6\cdot6=36\).
  6. \(4\sqrt7\cdot2\sqrt7=8\cdot7=56\).
  7. \((2\sqrt6-2\sqrt2+\sqrt8)\cdot4\sqrt2=(2\sqrt6-2\sqrt2+2\sqrt2)\cdot4\sqrt2\)
    \(=2\sqrt6\cdot4\sqrt2=8\sqrt{12}=16\sqrt3\).
  8. \((2\sqrt6-2\sqrt{54}+6\sqrt{96})\cdot2\sqrt3=(2\sqrt6-6\sqrt6+24\sqrt6)\cdot2\sqrt3\)
    \(=20\sqrt6\cdot2\sqrt3=40\sqrt{18}=120\sqrt2\).
  9. \((3\sqrt5-2\sqrt2)(3\sqrt5+2\sqrt2)=(3\sqrt5)^2-(2\sqrt2)^2=45-8=37\).
  10. \((7\sqrt3+2\sqrt6)(7\sqrt3-2\sqrt6)=(7\sqrt3)^2-(2\sqrt6)^2=147-24=123\).
  11. \((\sqrt5-\sqrt2)(\sqrt5+\sqrt2)=5-2=3\).
  12. \((\sqrt3-\sqrt6)(\sqrt3+\sqrt6)=3-6=-3\).

Ответ: 1) \(18\sqrt2-20\); 2) \(51\sqrt3\); 3) \(10\); 4) \(15\); 5) \(36\); 6) \(56\); 7) \(16\sqrt3\); 8) \(120\sqrt2\); 9) \(37\); 10) \(123\); 11) \(3\); 12) \(-3\).