If x = sqrt{frac{2+sqrt{3}}{2-sqrt{3}}},then | vedclass

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(D) Given $x = \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}$.Rationalizing the denominator inside the square root:$x = \sqrt{\frac{(2+\sqrt{3})(2+\sqrt{3})}{(

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(D) Given $x = \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}$.Rationalizing the denominator inside the square root:$x = \sqrt{\frac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}} = \sqrt{\frac{(2+\sqrt{3})^2}{4-3}} = \sqrt{(2+\sqrt{3})^2} = 2+\sqrt{3}$.Now,we need to find the value of $x^2(x-4)^2 = [x(x-4)]^2$.Substituting $x = 2+\sqrt{3}$:$x(x-4) = (2+\sqrt{3})(2+\sqrt{3}-4) = (2+\sqrt{3})(\sqrt{3}-2)$.Using the identity $(a+b)(a-b) = a^2-b^2$:$x(x-4) = (\sqrt{3}+2)(\sqrt{3}-2) = (\sqrt{3})^2 - (2)^2 = 3 - 4 = -1$.Therefore,$x^2(x-4)^2 = (-1)^2 = 1$.

If $x = \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}$,then $x^2(x-4)^2$ is equal to :

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