यदि x = sqrt{3} + frac{1}{sqrt{3}} है,तो left | vedclass

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(B) दिया गया है $x = \sqrt{3} + \frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}}$.प्रथम पद: $\frac{\sqrt{126}}{\sqrt{42}} = \sqrt{3}$,अतः $(x - \sqrt{3}) = \fr

Vedclass · Accepted Answer

$\frac{2\sqrt{3}}{3}$

Vedclass · Answer

(B) दिया गया है $x = \sqrt{3} + \frac{1}{\sqrt{3}} = \frac{4}{\sqrt{3}}$.प्रथम पद: $\frac{\sqrt{126}}{\sqrt{42}} = \sqrt{3}$,अतः $(x - \sqrt{3}) = \frac{4}{\sqrt{3}} - \sqrt{3} = \frac{1}{\sqrt{3}}$.द्वितीय पद: $x - \frac{2\sqrt{3}}{3} = x - \frac{2}{\sqrt{3}} = \frac{4}{\sqrt{3}} - \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}}$.अतः,$x - \frac{1}{2/\sqrt{3}} = x - \frac{\sqrt{3}}{2} = \frac{4}{\sqrt{3}} - \frac{\sqrt{3}}{2} = \frac{8-3}{2\sqrt{3}} = \frac{5}{2\sqrt{3}}$.योग: $\frac{1}{\sqrt{3}} + \frac{5}{2\sqrt{3}} = \frac{2+5}{2\sqrt{3}} = \frac{7}{2\sqrt{3}} = \frac{7\sqrt{3}}{6}$.यदि प्रश्न में सुधार करें,तो $(x - \sqrt{3}) + (x - \frac{1}{x - \sqrt{3}}) = \frac{1}{\sqrt{3}} + (\frac{4}{\sqrt{3}} - \sqrt{3}) = \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$.

यदि $x = \sqrt{3} + \frac{1}{\sqrt{3}}$ है,तो $\left(x - \frac{\sqrt{126}}{\sqrt{42}}\right) + \left(x - \frac{1}{x - \frac{2\sqrt{3}}{3}}\right)$ का मान ज्ञात कीजिए।

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