Câu hỏi của Muối Hóa Học

Question

Tính:$\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$

Lê Thị Thục Hiền · Accepted Answer

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

Nguyễn Lê Phước Thịnh · Answer

Ta có: $\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$$=\dfrac{\sqrt{2}-\sqrt{2+\sqrt{3}}}{2-2-\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{2-2+\sqrt{3}}$$=\dfrac{\sqrt{2}-\sqrt{2+\sqrt{3}}}{-\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{\sqrt{3}}$$=\dfrac{-\sqrt{2}-\sqrt{2+\sqrt{3}}}{\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{\sqrt{3}}$$=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{6}}$$=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{6}}$$=\dfrac{-2}{\sqrt{6}}=\dfrac{-2\sqrt{6}}{6}=\dfrac{-\sqrt{6}}{3}$Câu trả lời: Ta có: $\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$$=\dfrac{\sqrt{2}-\sqrt{2+\sqrt{3}}}{2-2-\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{2-2+\sqrt{3}}$$=\dfrac{\sqrt{2}-\sqrt{2+\sqrt{3}}}{-\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{\sqrt{3}}$$=\dfrac{-\sqrt{2}-\sqrt{2+\sqrt{3}}}{\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{\sqrt{3}}$$=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{6}}$$=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{6}}$$=\dfrac{-2}{\sqrt{6}}=\dfrac{-2\sqrt{6}}{6}=\dfrac{-\sqrt{6}}{3}$

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