Question

answer full solution

Answer 1

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

Let's simplify each term one by one,
$\frac{1}{ \sqrt{11 - 2 \sqrt{30} } } \\ here \\ let \sqrt{11 - 2 \sqrt{30} } = x + y \\ 11 - 2 \sqrt{30} = {(x - y)}^{2} \\ 11 - 2 \sqrt{30} = {x}^{2} + {y}^{2} - 2xy \\ \\ by \: hit \: and \: trial \: method \: \\ x = \sqrt{6} \\ y = \sqrt{5} \\ so \\ \sqrt{11 - 2 \sqrt{30} } = \sqrt{( {\sqrt{6} - \sqrt{5} ) }^{2} }$
by susbtitutingthis value for the very first term we're gonna get,
$\frac{1}{ \sqrt{11 - 2 \sqrt{30} } } = \frac{1}{ \sqrt{ { (\sqrt{6} - \sqrt{5} )}^{2} } } \\ \\ = \frac{1}{ \sqrt{6} - \sqrt{5} }$
Similarly, simply all other terms

$\frac{3}{ \sqrt{7 - 2 \sqrt{10} } } = \frac{3}{ \sqrt{ {( \sqrt{5 } - \sqrt{2} ) }^{2} } } = \frac{3}{ \sqrt{5} - \sqrt{2} }$

We'll do this for the last term too,

$\frac{4}{ \sqrt{8 + 4 \sqrt{3} } } = \frac{4}{ \sqrt{8 + 2 \sqrt{12} } } = \frac{4}{ \sqrt{ {( \sqrt{6} + \sqrt{2}) }^{2} } } \\ = \frac{4}{ \sqrt{6} + \sqrt{2} }$




Now, we have simplified all three terms
substitute them for the previous terms and you'll end up with:

$\frac{1}{ \sqrt{6} - \sqrt{5} } - \frac{3}{ \sqrt{5} - \sqrt{2} } - \frac{4}{ \sqrt{6} + \sqrt{2} }$




Rationalise each term one by one and you'll get

$\frac{ \sqrt{6} + \sqrt{5} }{1} - \frac{ 3(\sqrt{5} + \sqrt{2}) } {3} - \frac{ 4(\sqrt{6} - \sqrt{2} ) }{4} \\ = \sqrt{6} + \sqrt{5} - ( \sqrt{5} + \sqrt{2} ) - ( \sqrt{6} - \sqrt{2}) \\ = \sqrt{6} + \sqrt{5} - \sqrt{5} - \sqrt{2} - \sqrt{6} + \sqrt{2} \\ = 0$



Therefore, Option D is the correct one.

Answer 2

ans is D. sol is attached