a) \(=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{9-2}=\dfrac{6}{7}\)
b) \(=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{5-2\sqrt{15}+3+5+2\sqrt{15}+3}{5-3}=\dfrac{16}{2}=8\)
c) \(=\dfrac{2\left(3\sqrt{2}+4\right)-2\left(3\sqrt{2}-4\right)}{\left(3\sqrt{2}-4\right)\left(3\sqrt{2}+4\right)}=\dfrac{6\sqrt{2}+8-6\sqrt{2}+8}{18-16}=\dfrac{16}{2}=8\)
d) \(=\dfrac{3\left(2\sqrt{2}+3\sqrt{3}\right)-3\left(2\sqrt{2}-3\sqrt{3}\right)}{\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)}=\dfrac{6\sqrt{2}+9\sqrt{3}-6\sqrt{2}+9\sqrt{3}}{8-27}=\dfrac{18\sqrt{3}}{-19}=-\dfrac{18\sqrt{3}}{19}\)
\(a,=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}=\dfrac{6}{3^2-\left(\sqrt{2}\right)^2}=\dfrac{6}{7}\\ b,=\dfrac{2\left(3\sqrt{2}+4\right)-2\left(3\sqrt{2}-4\right)}{\left(3\sqrt{2}-4\right)\left(3\sqrt{2}+4\right)}=\dfrac{6\sqrt{2}+8-6\sqrt{2}+8}{18-16}=\dfrac{16}{2}=8\\ c,=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}=\dfrac{16}{2}=8\)
\(d,=\dfrac{3\left(2\sqrt{2}+3\sqrt{3}\right)}{8-27}-\dfrac{3}{5\sqrt{3}}=\dfrac{6\sqrt{2}+9\sqrt{3}}{-19}-\dfrac{\sqrt{3}}{5}\\ =\dfrac{-30\sqrt{2}-45\sqrt{3}-19\sqrt{3}}{95}=\dfrac{-30\sqrt{2}-64\sqrt{2}}{95}\)
chỉ cần lm B1 thôi nhé