Question

. Please do this I need to submit

Answer 1

Hey friend,
Here is the answer you were looking for:
$(i) \frac{1}{ \sqrt{8} }$

On rationalizing the denominator we get,

$= \frac{1}{ \sqrt{8} } \times \frac{ \sqrt{8} }{ \sqrt{8} } \\ \\ = \frac{ \sqrt{8} }{ {( \sqrt{8} )}^{2} } \\ \\ = \frac{ \sqrt{8} }{8}$

$(ii) \frac{ \sqrt{5} }{2 \sqrt{3} }$

On rationalizing the denominator we get,

$= \frac{ \sqrt{5} }{2 \sqrt{3} } \times \frac{ 2\sqrt{3} }{2 \sqrt{3} } \\ \\ = \frac{ \sqrt{5} \times 2 \sqrt{3} }{ {(2 \sqrt{3} )}^{2} } \\ \\ = \frac{2 \sqrt{15} }{12} \\ \\ = \frac{ \sqrt{15} }{6}$

$(iii) \frac{1}{4 + \sqrt{3} }$

On rationalizing the denominator we get,

$= \frac{1}{4 + \sqrt{3} } \times \frac{4 - \sqrt{3} }{4 - \sqrt{3} }$

Using the denominator we get,

$(x + y)(x - y) = {x}^{2} - {y}^{2}$

$= \frac{ 4 - \sqrt{3} }{ {(4)}^{2} - {( \sqrt{3} )}^{2} } \\ \\ = \frac{4 - \sqrt{3} }{16 - 3} \\ \\ = \frac{4 - \sqrt{3} }{13}$
$(iv) \frac{11 - 2 \sqrt{2} }{11 + 2 \sqrt{2} }$

On rationalizing the denominator we get,

$= \frac{11 - 2 \sqrt{2} }{11 + 2 \sqrt{2} } \times \frac{11 - 2 \sqrt{2} }{11 - 2 \sqrt{2} }$

Using the identities :

$= \frac{ {(11)}^{2} + {(2 \sqrt{2} )}^{2} - 2(11)(2 \sqrt{2} )}{ {(11)}^{2} - {(2 \sqrt{2} )}^{2} } \\ \\ = \frac{121 + 8 - 44 \sqrt{2} }{121 - 8} \\ \\ = \frac{129 - 44 \sqrt{2} }{113}$

Hope this helps!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
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