left(2+sqrt{2}+frac{1}{2+sqrt{2}}+frac{1}{sqr | vedclass

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(B) આપેલ પદાવલિ: $2+\sqrt{2}+\frac{1}{2+\sqrt{2}}+\frac{1}{\sqrt{2}-2}$છેદનું સંમેયીકરણ કરતા:$\frac{1}{2+\sqrt{2}} = \frac{1}{2+\sqrt{2}} 	imes \frac

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$2$

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(B) આપેલ પદાવલિ: $2+\sqrt{2}+\frac{1}{2+\sqrt{2}}+\frac{1}{\sqrt{2}-2}$છેદનું સંમેયીકરણ કરતા:$\frac{1}{2+\sqrt{2}} = \frac{1}{2+\sqrt{2}} 	imes \frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{2-\sqrt{2}}{4-2} = \frac{2-\sqrt{2}}{2} = 1 - \frac{\sqrt{2}}{2}$$\frac{1}{\sqrt{2}-2} = \frac{1}{\sqrt{2}-2} 	imes \frac{\sqrt{2}+2}{\sqrt{2}+2} = \frac{\sqrt{2}+2}{2-4} = \frac{\sqrt{2}+2}{-2} = -\frac{\sqrt{2}}{2} - 1$આ કિંમતોને મૂળ પદાવલિમાં મુકતા:$= 2 + \sqrt{2} + (1 - \frac{\sqrt{2}}{2}) + (-1 - \frac{\sqrt{2}}{2})$$= 2 + \sqrt{2} + 1 - \frac{\sqrt{2}}{2} - 1 - \frac{\sqrt{2}}{2}$$= 2 + (1 - 1) + (\sqrt{2} - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2})$$= 2 + 0 + (\sqrt{2} - \sqrt{2}) = 2$

$\left(2+\sqrt{2}+\frac{1}{2+\sqrt{2}}+\frac{1}{\sqrt{2}-2}\right) = ?$

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