Lời giải:
a) Đặt \(\sqrt{3+\sqrt{2}}=a; \sqrt{3-\sqrt{2}}=b\Rightarrow \left\{\begin{matrix}
a^2+b^2=6\\
a^2-b^2=2\sqrt{2}\\
ab=\sqrt{(3+\sqrt{2})(3-\sqrt{2})}=\sqrt{7}\end{matrix}\right.\)
\(A=\frac{a+b}{a-b}=\frac{(a+b)^2}{(a+b)(a-b)}=\frac{(a+b)^2}{a^2-b^2}=\frac{a^2+b^2+2ab}{a^2-b^2}\)
\(=\frac{6+2\sqrt{7}}{2\sqrt{2}}=\frac{3+\sqrt{7}}{\sqrt{2}}\)
b)
Đặt \(\sqrt{5+3\sqrt{2}}=a; \sqrt{5-3\sqrt{2}}=b\)
\(\Rightarrow \left\{\begin{matrix} a^2+b^2=10\\ a^2-b^2=6\sqrt{2}\\ ab=\sqrt{(5+3\sqrt{2})(5-3\sqrt{2})}=\sqrt{25-(3\sqrt{2})^2}=\sqrt{7}\end{matrix}\right.\)
\(B=\frac{\sqrt{9(5+3\sqrt{2})}+\sqrt{9(5-3\sqrt{2})}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}=\frac{3(a+b)}{a-b}\)
\(=\frac{3(a+b)^2}{a^2-b^2}=\frac{3(a^2+b^2+2ab)}{a^2-b^2}=\frac{3(10+2\sqrt{7})}{6\sqrt{2}}\)
\(=\frac{5+\sqrt{7}}{\sqrt{2}}\)