Question

Answer 1

`a)`\(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}=\dfrac{6}{7}\)

`b)`\(\dfrac{2}{3\sqrt{2}-4}-\dfrac{2}{3\sqrt{2}+4}=2.\left(\dfrac{3\sqrt{2}+4-3\sqrt{2}+4}{\left(3\sqrt{2}-4\right)\left(3\sqrt{2}+4\right)}\right)=2.\dfrac{8}{2}=8\)

`c)`\(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}=\dfrac{16}{2}=8\)