Question

\(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2}{\sqrt{3}-1}\)

Answer 1

\(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2}{\sqrt{3}-1}=\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{3}+\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+\dfrac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\dfrac{\sqrt{2}}{2}+\dfrac{2\sqrt{3}+2}{3-1}-\dfrac{\sqrt{3}+\sqrt{2}}{3-2}=\dfrac{\sqrt{2}+2\sqrt{3}+2}{2}-\left(\sqrt{3}+\sqrt{2}\right)=\dfrac{\sqrt{2}+2\sqrt{3}+2-2\left(\sqrt{3}+\sqrt{2}\right)}{2}=\dfrac{\sqrt{2}+2\sqrt{3}+2-2\sqrt{3}-2\sqrt{2}}{2}=\dfrac{2-\sqrt{2}}{2}\)