Question

\(\dfrac{3}{\sqrt{5}+\sqrt{2}}\) + \(\dfrac{1}{\sqrt{2}-1}\) - \(\dfrac{4}{3-\sqrt{5}}\). Rút Gọn

Answer 1

\(\dfrac{3}{\sqrt{5}+\sqrt{2}}+\dfrac{1}{\sqrt{2}-1}-\dfrac{4}{3-\sqrt{5}}\)

\(=\dfrac{3\left(\sqrt{2}-1\right)\left(3-\sqrt{5}\right)}{\left(\sqrt{2}-1\right)\left(3-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}+\dfrac{3\sqrt{5}+3\sqrt{2}-5-\sqrt{10}}{\left(\sqrt{2}-1\right)\left(3-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)}-\dfrac{4\left(\sqrt{10}+2-\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{2}-1\right)\left(3-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)}\)

\(=\dfrac{16\sqrt{2}-8\sqrt{10}-22+10\sqrt{5}}{\left(\sqrt{2}-1\right)\left(3-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}=\dfrac{16\sqrt{2}-8\sqrt{10}-22+10\sqrt{5}}{4\sqrt{10}-8\sqrt{2}+11-5\sqrt{5}}\)

\(=-2.\)