Respuesta :

Answer:

[tex]{ \tt{ {x}^{2} - \frac{1}{ {x}^{2} } }} \\ = { \tt{ {(2 + \sqrt{5} )}^{2} - \frac{1}{ {(2 + \sqrt{5}) }^{2} } }} \\ = { \tt{ \frac{(2 + \sqrt{5} ) {}^{4} - 1}{ {(2 + \sqrt{5} )}^{2} } }} \\ = { \tt{ \frac{(9 + 4 \sqrt{5}) {}^{2} }{ {(9 + 4\sqrt{5}) }}}} \\ = { \tt{9 + 4 \sqrt{5} }}[/tex]

devishri1977

Answer:

[tex]8\sqrt{5}[/tex]

Step-by-step explanation:

[tex]x = 2 + \sqrt{5}\\\\ x^{2} = (2+ \sqrt{5})^{2} \\\\ \ \ \ \ = 2^{2}+2* \sqrt{5}*2+( \sqrt{5})^{2}\\\\[/tex]

 [tex]= 4 + 4 \sqrt{5}+5\\\\= 9+4 \sqrt{5}[/tex]

[tex]\frac{1}{x^{2}}=\frac{1}{9+4\sqrt{5}}\\\\=\frac{1*(9-4\sqrt{5}}{(9+4\sqrt{5})(9-4\sqrt{5})}\\\\=\frac{9-4\sqrt{5}}{9^{2}-(4\sqrt{5})^{2}}\\\\=\frac{9-4\sqrt{5}}{81-4^{2}(\sqrt{5})^{2}}\\\\=\frac{9-4\sqrt{5}}{81-16*5}\\\\=\frac{9-4\sqrt{5}}{81-80}\\\\=\frac{9-4\sqrt{5}}{1}\\\\=9-4\sqrt{5}[/tex]

[tex]x^{2}-\frac{1}{x^{2}}= 9 + 4\sqrt{5} -(9 - 4\sqrt{5})\\\\[/tex]

            [tex]= 9 + 4\sqrt{5} - 9 + 4\sqrt{5}\\\\= 9 - 9 + 4\sqrt{5} + 4\sqrt{5}\\\\= 8\sqrt{5}[/tex]

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