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\(y=\frac{1}{9+4\sqrt{5}}=\frac{1}{\left(\sqrt{5}+2\right)^2}\)
\(\Rightarrow N=\frac{1}{\left(\sqrt{5}-2\right)^2}-\frac{3}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}+\frac{2}{9+4\sqrt{5}}\)
\(=\frac{1}{9-4\sqrt{5}}+\frac{2}{9+4\sqrt{5}}-3=\frac{9+4\sqrt{5}+18-8\sqrt{5}}{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}-3=24-4\sqrt{5}\)
\(S^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=x^2+y^2+x^2y^2+1+x^2y^2-1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
\(=\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+x^2y^2-1\)
\(=\left(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right)^2-1\)
\(=2005^2-1\)
\(\Rightarrow S=\pm\sqrt{2005^2-1}\)
c/
Giả sử \(\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}< 2\sqrt[3]{3}\)
\(\Leftrightarrow\sqrt[3]{3+\sqrt[3]{3}}-\sqrt[3]{3}< \sqrt[3]{3}-\sqrt[3]{3-\sqrt[3]{3}}\)
\(\Leftrightarrow\frac{\sqrt[3]{3}}{\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}+\sqrt[3]{9+3\sqrt[3]{3}}+\sqrt[3]{9}}< \frac{\sqrt[3]{3}}{\sqrt[3]{9}+\sqrt[3]{9-3\sqrt[3]{3}}+\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}}\)
\(\Leftrightarrow\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}+\sqrt[3]{9+3\sqrt[3]{3}}+\sqrt[3]{9}>\sqrt[3]{9}+\sqrt[3]{9-3\sqrt[3]{3}}+\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}\)
\(\Leftrightarrow\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}+\sqrt[3]{9+3\sqrt[3]{3}}>\sqrt[3]{9-3\sqrt[3]{3}}+\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}\) (1)
Ta có: \(\left\{{}\begin{matrix}\sqrt[3]{9+3\sqrt[3]{3}}>\sqrt[3]{9-3\sqrt[3]{3}}\\\sqrt[3]{\left(3+\sqrt[3]{3}\right)^2}>\sqrt[3]{\left(3-\sqrt[3]{3}\right)^2}\end{matrix}\right.\)
Nên (1) đúng
Vậy BĐT ban đầu đúng