Question

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

Answer 1

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

Answer 2

d) \(A=\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)

\(\Rightarrow A^2=\dfrac{2-\sqrt{3}}{2+\sqrt{3}}+2\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}\cdot\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}+\dfrac{2+\sqrt{3}}{2-\sqrt{3}}\)

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

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

          \(=\dfrac{14}{1}+2=16\)  

\(\Rightarrow A=\sqrt{16}=4\) (KHÔNG THỂ BẰNG -4 vì biểu thức này luôn có nghĩa tức luôn xác định)