જો x = 3 - sqrt{5} હોય,તો frac{sqrt{x}}{sqrt{ | vedclass

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Question

(D) આપેલ છે $x = 3 - \sqrt{5}$.પ્રથમ,$\sqrt{x} = \sqrt{3 - \sqrt{5}} = \frac{1}{\sqrt{2}} \sqrt{6 - 2\sqrt{5}} = \frac{1}{\sqrt{2}}(\sqrt{5} - 1)$.ત્ય

Vedclass · Accepted Answer

$1/\sqrt{5}$

Vedclass · Answer

(D) આપેલ છે $x = 3 - \sqrt{5}$.પ્રથમ,$\sqrt{x} = \sqrt{3 - \sqrt{5}} = \frac{1}{\sqrt{2}} \sqrt{6 - 2\sqrt{5}} = \frac{1}{\sqrt{2}}(\sqrt{5} - 1)$.ત્યારબાદ,$3x - 2 = 3(3 - \sqrt{5}) - 2 = 9 - 3\sqrt{5} - 2 = 7 - 3\sqrt{5}$.$\sqrt{3x - 2}$ ને સરળ બનાવવા માટે,$7 - 3\sqrt{5} = \frac{14 - 6\sqrt{5}}{2} = \frac{(3 - \sqrt{5})^2}{2}$.તેથી,$\sqrt{3x - 2} = \frac{3 - \sqrt{5}}{\sqrt{2}}$.હવે,$\sqrt{2} + \sqrt{3x - 2} = \sqrt{2} + \frac{3 - \sqrt{5}}{\sqrt{2}} = \frac{2 + 3 - \sqrt{5}}{\sqrt{2}} = \frac{5 - \sqrt{5}}{\sqrt{2}} = \frac{\sqrt{5}(\sqrt{5} - 1)}{\sqrt{2}}$.અંતે,$\frac{\sqrt{x}}{\sqrt{2} + \sqrt{3x - 2}} = \frac{\frac{1}{\sqrt{2}}(\sqrt{5} - 1)}{\frac{\sqrt{5}}{\sqrt{2}}(\sqrt{5} - 1)} = \frac{1}{\sqrt{5}}$.

જો $x = 3 - \sqrt{5}$ હોય,તો $\frac{\sqrt{x}}{\sqrt{2} + \sqrt{3x - 2}} = $

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