Câu hỏi của Thành Nam

Question

$\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$

Nguyễn Việt Lâm · Accepted Answer

$=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}$$=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{3+\sqrt{3}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\sqrt{2}\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}$$=\dfrac{6\sqrt{2}}{6}=\sqrt{2}$Câu trả lời: $=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}$$=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{3+\sqrt{3}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\sqrt{2}\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}$$=\dfrac{6\sqrt{2}}{6}=\sqrt{2}$

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