[tex]\it a)\ \ A=\sqrt{9\cdot2}\cdot\sqrt3(\sqrt{49}-\sqrt{36}+\sqrt{100}-\sqrt{144})=3\sqrt6(7-6+10-12)=\\ \\ =3\sqrt6\cdot(-1)=-3\sqrt6\\ \\ \\ B=\sqrt{16\cdot3}\cdot\sqrt{2}(\sqrt{25}-\sqrt{16}+\sqrt{64}-\sqrt{49})=4\sqrt6(5-4+8-7)=\\ \\ =4\sqrt6\cdot2=8\sqrt6[/tex]
[tex]\it A\cdot B=-3\sqrt6\cdot8\sqrt6=-24\cdot6=-144\\ \\ \dfrac{A}{B}=\dfrac{-3\sqrt6}{8\sqrt6}=-\dfrac{3}{8}[/tex]
