Question

bài 1 : rút gọn biểu thức

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

Answer 1

nhan them \(\sqrt{2}\)

vao mau voi tu la dc

Answer 2

Đặt \(x=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2;y=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\) thì \(3+\sqrt{5}=\dfrac{x}{2};3-\sqrt{5}=\dfrac{y}{2}\) nên nếu kí hiệu biểu thức đã cho là A thì ta có

\(A=\dfrac{\dfrac{x}{2}}{2\sqrt{2}+\sqrt{\dfrac{x}{2}}}+\dfrac{\dfrac{y}{2}}{2\sqrt{2}-\sqrt{\dfrac{y}{2}}}=\dfrac{x}{4\sqrt{2}+\sqrt{2}\sqrt{x}}+\dfrac{y}{4\sqrt{2}-\sqrt{2}\sqrt{y}}\)

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

\(=\dfrac{\left(\sqrt{5}+1\right)^2}{\sqrt{2}\sqrt{5}\left(\sqrt{5}+1\right)}+\dfrac{\left(\sqrt{5}-1\right)^2}{\sqrt{2}\sqrt{5}\left(\sqrt{5}-1\right)}\)

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

\(=\sqrt{2}\)