Question

Bài1: Tính

a) \(\sqrt{4-\sqrt{15}}+\sqrt{2-\sqrt{3}}\)

b) \(\sqrt{4-\sqrt{7}}+\sqrt{8-3\sqrt{7}}\)

Answer 1

a. \(\sqrt{4-\sqrt{15}}+\sqrt{2-\sqrt{3}}=\sqrt{\left(\sqrt{\dfrac{5}{2}}-\sqrt{\dfrac{3}{2}}\right)^2}+\sqrt{\left(\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{1}{2}}\right)^2}=\sqrt[]{\dfrac{5}{2}}-\sqrt{\dfrac{3}{2}}+\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{1}{2}}=\dfrac{\sqrt{5}-1}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{2}}{2}\)

b. \(\sqrt{4-\sqrt{7}}+\sqrt{8-3\sqrt{7}}=\sqrt{\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{1}{2}}\right)^2}+\sqrt{\left(\sqrt{\dfrac{9}{2}}-\sqrt{\dfrac{7}{2}}\right)^2}=\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{1}{2}}+\sqrt{\dfrac{9}{2}}-\sqrt{\dfrac{7}{2}}=\dfrac{3-1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)