\(F=\left(\dfrac{1}{3-\sqrt{5}}+\dfrac{1}{3+\sqrt{5}}\right):\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\dfrac{6}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}:\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}=\dfrac{3}{2}.\dfrac{1}{\sqrt{5}}=\dfrac{3}{2\sqrt{5}}\)
\(G=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}=\dfrac{\sqrt{5+2\sqrt{5}+1}+\sqrt{9-2.3.\sqrt{5}+5}-2}{\sqrt{2}}=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
\(H=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{x-2+2\sqrt{2}.\sqrt{x-2}+2}+\sqrt{x-2-2\sqrt{2}.\sqrt{x-2}+2}=\sqrt{\left(\sqrt{x-2}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{x-2}-\sqrt{2}\right)^2}=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|\left(x\ge2\right)\)
+) ta có : \(F=\left(\dfrac{1}{3-\sqrt{5}}+\dfrac{1}{3+\sqrt{5}}\right):\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\)
\(\Leftrightarrow F=\left(\dfrac{3+\sqrt{5}+3-\sqrt{5}}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right):\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\)
\(\Leftrightarrow F=\dfrac{6}{4}.\dfrac{1}{\sqrt{5}}=\dfrac{6}{4\sqrt{5}}\)
+) ta có : \(G=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(\Leftrightarrow G=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{2}}\)
\(\Leftrightarrow G=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}-2}{\sqrt{2}}\)
\(\Leftrightarrow G=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
+) ta có : \(H=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)
\(\Rightarrow H^2=2x+2\sqrt{x^2-4\left(2x-4\right)}=2x+2\sqrt{\left(x-4\right)^2}\)
\(\Leftrightarrow\left[{}\begin{matrix}H^2=4x-8\\H^2=8\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}H=\sqrt{4x-8}\\H=-\sqrt{4x-8}\left(L\right)\end{matrix}\right.\\\left[{}\begin{matrix}H=\sqrt{8}\\H=-\sqrt{8}\left(L\right)\end{matrix}\right.\end{matrix}\right.\)
vậy \(H=\sqrt{4x-8};H=\sqrt{8}\)
