\(A=\dfrac{3\sqrt{2^2.2}-2\sqrt{2^2.3}+\sqrt{2^2.5}}{3\sqrt{3^2.2}-2\sqrt{3^3.3}+\sqrt{3^2}.5}=\dfrac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)
\(=\dfrac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}=\dfrac{2}{3}\)
\(B=\dfrac{2\sqrt{3.5}-2\sqrt{2.5}+\sqrt{2.3}-3}{2\sqrt{5}-2\sqrt{2.5}-\sqrt{3}+\sqrt{2.3}}=\dfrac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}\left(1-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}\)
\(=\dfrac{\left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{5}-\sqrt{3}\right)}{\left(1-\sqrt{2}\right)\left(2\sqrt{5}-\sqrt{3}\right)}=\dfrac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)
