Question

Tính:

\(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-\sqrt{3-\sqrt{5}}\)

Answer 1

Ta có: 4 + √15 = 2(4 + √15) / 2 = 8 + 2√15 / 2 = (√5 + √3)² /2 = [(√5 + √3)/√2]²

4 - √15 = 2(4 - √15) /2 = 8 - 2√15 /2 = (√5 - √3) ² /2 =[(√5 - √3)/√2]²

3 - √5 = 2(3 - √5) /2 = 6 - 2√5 /2 = (√5 - 1)² /2 = [(√5 - 1)√2]²

Thay vào biểu thức ban đầu ta được kết quả = 3√10 - √2 / 2

Answer 2

\(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-\sqrt{3-\sqrt{5}}=\dfrac{\sqrt{2}.\left(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-\sqrt{3-\sqrt{5}}\right)}{\sqrt{2}}=\dfrac{\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\dfrac{\sqrt{5+2\sqrt{5}.\sqrt{3}+3}+\sqrt{5-2\sqrt{5}.\sqrt{3}+3}-\sqrt{5-2\sqrt{5}+1}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\dfrac{\left|\sqrt{5}+\sqrt{3}\right|+\left|\sqrt{5}-\sqrt{3}\right|-\left|\sqrt{5}-1\right|}{\sqrt{2}}=\dfrac{\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-\sqrt{5}+1}{\sqrt{2}}=\dfrac{\sqrt{5}+1}{\sqrt{2}}=\dfrac{\sqrt{10}+\sqrt{2}}{2}\)