Đáp án:
$F=\sqrt{2}$
Giải thích các bước giải:
$F=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}\\=\sqrt{\dfrac{6+2\sqrt{5}}{2}}+\sqrt{\dfrac{14-6\sqrt{5}}{2}}-\sqrt{2}\\=\dfrac{\sqrt{(\sqrt{5}+1)^2}}{\sqrt{2}}+\dfrac{\sqrt{(3-\sqrt{5})^2}}{\sqrt{2}}-\sqrt{2}\\=\dfrac{|\sqrt{5}+1|}{\sqrt{2}}+\dfrac{|3-\sqrt{5}|}{\sqrt{2}}-\sqrt{2}\\=\dfrac{\sqrt{5}+1+3-\sqrt{5}}{\sqrt{2}}-\sqrt{2}\\=\dfrac{4}{\sqrt{2}}-\sqrt{2}\\=\dfrac{4-2}{\sqrt{2}}\\=\sqrt{2}$
