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So sánh hai biểu thức sau x=2+căn3/căn2+căn(2+căn3) + 2−căn3/căn2−căn(2−căn3)

Hãy so sánh hai biểu thức sau:

\(x=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

\(y=\dfrac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\dfrac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\)

Theo dõi Vi phạm
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Trả lời (1)

  • nguyen thu

    \(x=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{3+1+2\sqrt{3}}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{3+1-2\sqrt{3}}}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{3}+1}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{3+\sqrt{3}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}\\ =\dfrac{2\sqrt{2}+\sqrt{6}}{3+\sqrt{3}}+\dfrac{2\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}\\ =\dfrac{\left(2\sqrt{2}+\sqrt{6}\right)\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\dfrac{\left(2\sqrt{2}-\sqrt{6}\right)\left(\sqrt{3}+1\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\\ =\dfrac{2\sqrt{6}+3\sqrt{2}-2\sqrt{2}-\sqrt{6}+2\sqrt{6}-3\sqrt{2}+2\sqrt{2}-\sqrt{6}}{2\sqrt{3}}\\ =\dfrac{4\sqrt{6}-2\sqrt{6}}{2\sqrt{3}}=\dfrac{\sqrt{6}\left(4-2\right)}{2\sqrt{3}}=\dfrac{2\sqrt{6}}{2\sqrt{3}}=\sqrt{2}\)

    \(y=\dfrac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\dfrac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\\ =\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{5}+\sqrt{6+2\sqrt{5}}}-\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{5}+\sqrt{6-2\sqrt{5}}}\\ =\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{5}+\sqrt{5+1+2\sqrt{5}}}-\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{5}+\sqrt{5+1-2\sqrt{5}}}\\ =\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{5}+\sqrt{\left(\sqrt{5}+1\right)^2}}-\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{5}+\sqrt{5}+1}-\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{5}+\sqrt{5}-1}\\ =\dfrac{3\sqrt{2}+\sqrt{10}}{\sqrt{5}+1}-\dfrac{3\sqrt{2}-\sqrt{10}}{\sqrt{5}-1}\\ =\dfrac{\left(3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}-\dfrac{\left(3\sqrt{2}-\sqrt{10}\right)\left(\sqrt{5}+1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\\ =\dfrac{3\sqrt{10}+5\sqrt{2}-3\sqrt{2}-\sqrt{10}-3\sqrt{10}+5\sqrt{2}-3\sqrt{2}+\sqrt{10}}{4}\\ =\dfrac{4\sqrt{2}}{4}=\sqrt{2}\)

    Vậy \(x=y\)

      bởi nguyen thu 24/01/2019
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