Đáp án:
`#DDD`
$\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\dfrac{\sqrt{3}.\sqrt{2}+\sqrt{7}.\sqrt{2}}{2\sqrt{3}+2\sqrt{7}}\\=\dfrac{\sqrt{2}(\sqrt{3}+\sqrt{7})}{2(\sqrt{3}+\sqrt{7})}=\dfrac{1}{\sqrt{2}}$
$\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\\=\dfrac{\sqrt{2}+\sqrt{3}+2+\sqrt{2}.\sqrt{3}+2\sqrt{2}+2}{\sqrt{2}+\sqrt{3}+2}\\=\dfrac{\sqrt{2}+\sqrt{3}+2+\sqrt{2}(\sqrt{2}+\sqrt{3}+2)}{\sqrt{2}+\sqrt{3}+2}\\=\dfrac{(1+\sqrt{2})(\sqrt{2}+\sqrt{3}+2)}{\sqrt{2}+\sqrt{3}+2}\\=1+\sqrt{2}$
Giải thích các bước giải:
