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If $x=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}$, then $x^2(x-4)^2$ is equal to :
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1
$\because \quad x=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}=\sqrt{\frac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}}$
$=\frac{2+\sqrt{3}}{\sqrt{4-3}}=2+\sqrt{3}$
$\therefore x^2(x-4)^2=(2+\sqrt{3})^2(2+\sqrt{3}-4)^2$
$=(\sqrt{3}+2)^2(\sqrt{3}-2)^2$
$=\left[(\sqrt{3})^2-(2)^2\right]^2$
$=(3-4)^2=(-1)^2$
$=1$
$=\frac{2+\sqrt{3}}{\sqrt{4-3}}=2+\sqrt{3}$
$\therefore x^2(x-4)^2=(2+\sqrt{3})^2(2+\sqrt{3}-4)^2$
$=(\sqrt{3}+2)^2(\sqrt{3}-2)^2$
$=\left[(\sqrt{3})^2-(2)^2\right]^2$
$=(3-4)^2=(-1)^2$
$=1$
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