Respuesta :

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[tex]\displaystyle \\ \frac{ \sqrt{2} -\sqrt{6}}{\sqrt{2} +\sqrt{6}} = \frac{ (\sqrt{2} -\sqrt{6})(\sqrt{2} -\sqrt{6})}{(\sqrt{2} +\sqrt{6})(\sqrt{2} -\sqrt{6})} = \\ \\ \\ = \frac{ (\sqrt{2} -\sqrt{6})^2}{(\sqrt{2})^2 -(\sqrt{6})^2} = \frac{ 2-2\sqrt{2}\sqrt{6} + 6}{2 -6} = \\ \\ \\ = \frac{ 8-2\sqrt{12} }{-4} = \frac{ 8-2\sqrt{4\times 3} }{-4} = \frac{ 8-4\sqrt{3} }{-4} = \boxed{-2+\sqrt{3}}[/tex]



calculista

Answer:

[tex]-2+\sqrt{3}[/tex]

Step-by-step explanation:

we have

[tex]\frac{\sqrt{2}-\sqrt{6}}{\sqrt{2}+\sqrt{6}}[/tex]

Multiply the expression by the conjugate of denominator

[tex]\frac{\sqrt{2}-\sqrt{6}}{\sqrt{2}+\sqrt{6}}*\frac{\sqrt{2}-\sqrt{6}}{\sqrt{2}-\sqrt{6}}\\ \\=-(\sqrt{2}-\sqrt{6})^{2}/4[/tex]  

[tex]=-(1/4)*(2-4\sqrt{3}+6)\\ \\=-(1/4)*(8-4\sqrt{3})\\ \\=-2+\sqrt{3}[/tex]

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