frac{1}{sqrt{11 - 2sqrt{30}}} - frac{3}{sqrt{ | vedclass

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(A) First,simplify each radical expression in the denominator:$1$. $\sqrt{11 - 2\sqrt{30}} = \sqrt{(\sqrt{6})^2 + (\sqrt{5})^2 - 2\sqrt{6}\sqrt{5}} =

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(A) First,simplify each radical expression in the denominator:$1$. $\sqrt{11 - 2\sqrt{30}} = \sqrt{(\sqrt{6})^2 + (\sqrt{5})^2 - 2\sqrt{6}\sqrt{5}} = \sqrt{(\sqrt{6} - \sqrt{5})^2} = \sqrt{6} - \sqrt{5}$.So,$\frac{1}{\sqrt{6} - \sqrt{5}} = \sqrt{6} + \sqrt{5}$.$2$. $\sqrt{7 - 2\sqrt{10}} = \sqrt{(\sqrt{5})^2 + (\sqrt{2})^2 - 2\sqrt{5}\sqrt{2}} = \sqrt{(\sqrt{5} - \sqrt{2})^2} = \sqrt{5} - \sqrt{2}$.So,$\frac{3}{\sqrt{5} - \sqrt{2}} = \frac{3(\sqrt{5} + \sqrt{2})}{5 - 2} = \frac{3(\sqrt{5} + \sqrt{2})}{3} = \sqrt{5} + \sqrt{2}$.$3$. $\sqrt{8 + 4\sqrt{3}} = \sqrt{8 + 2\sqrt{12}} = \sqrt{(\sqrt{6})^2 + (\sqrt{2})^2 + 2\sqrt{6}\sqrt{2}} = \sqrt{(\sqrt{6} + \sqrt{2})^2} = \sqrt{6} + \sqrt{2}$.So,$\frac{4}{\sqrt{6} + \sqrt{2}} = \frac{4(\sqrt{6} - \sqrt{2})}{6 - 2} = \frac{4(\sqrt{6} - \sqrt{2})}{4} = \sqrt{6} - \sqrt{2}$.Now,substitute these values back into the expression:$(\sqrt{6} + \sqrt{5}) - (\sqrt{5} + \sqrt{2}) - (\sqrt{6} - \sqrt{2})$$= \sqrt{6} + \sqrt{5} - \sqrt{5} - \sqrt{2} - \sqrt{6} + \sqrt{2} = 0$.

$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y + 5 = 0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x = 1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x + 8y = 27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x = 9$

$\frac{1}{\sqrt{11 - 2\sqrt{30}}} - \frac{3}{\sqrt{7 - 2\sqrt{10}}} - \frac{4}{\sqrt{8 + 4\sqrt{3}}} = $

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