Question

________________________
$\maltese \: \bold{rationalise \: the \: denominator \: and \: simplify :}$
$\sf \: 1. \: \frac{4 + \sqrt{5} }{4 - \sqrt{5} } + \frac{4 - \sqrt{5} }{4 + \sqrt{5} }$
$\sf \: 2. \: \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$
$\sf3. \: \frac{3}{5 - \sqrt{3} } + \frac{2}{5 + \sqrt{3} }$
$\huge \pmb{ \frak{note : - }}$
$\bold{ \leadsto \: give \: right \: answer \: only}$
$\bold{ \leadsto \: do \: not \: spam }$
$\bold {\leadsto \: questions \: are \: available \: in \: attachment \: too}$
________________________​

Answer 1

Answer:

this is the answer. however just cross check once if there is any calculation error.

hope it helps

Answer 2

Answer:

You will get it below

And can you also tell from where did you got these questions ?

Step-by-step explanation:

1)

$\frac{4 + \sqrt{5} }{4 - \sqrt{5} } + \frac{4 - \sqrt{5} }{4 + \sqrt{5} }$

Rationalizing the denominator-

$[tex] \frac{4 + \sqrt{5} }{4 - \sqrt{5} } \times \frac{4 + \sqrt{5} }{4 + \sqrt{5} } \\ \\ = > \frac{(4 + \sqrt{5} )^{2} }{(4)^{2} - ( { \sqrt{5} })^{2} } \\ \\ = > \frac{16 + 8 \sqrt{5} + 5 }{16 - 5} \\ \\ = > \frac{21 + 8 \sqrt{5} }{11} \\ \\ \frac{4 - \sqrt{5} }{4 + \sqrt{5} } \times \frac{4 - \sqrt{5} }{4 - \sqrt{5} } \\ \\ = > \frac{(4 - \sqrt{5} )^{2} }{ ({4})^{2} - {( \sqrt{5} )}^{2} } \\ \\ = > \frac{16 - 8 \sqrt{5} + 5}{16 - 5} \\ \\ = > \frac{21 - 8 \sqrt{5} }{11}$[/tex]

Therefore,

$\frac{4 + \sqrt{5} }{4 - \sqrt{5} } + \frac{4 - \sqrt{5} }{4 + \sqrt{5} } \\ = \frac{21 + 8 \sqrt{5} }{11} + \frac{21 - 8 \sqrt{5} }{11} \\ \\ = > \frac{(21 + 8 \sqrt{5}) + (21 - 8 \sqrt{5}) }{11} \\ \\ = > \frac{21 + 8 \sqrt{5} + 21 - 8 \sqrt{5} }{11} \\ \\ = > \frac{42}{11}$

2)

$\frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3 - \sqrt{2} } } \\ \\ = > \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \\ = > \frac{( \sqrt{3} + \sqrt{2})^{2} }{( \sqrt{3})^{2} - ({ \sqrt{2} )}^{2} } \\ \\ = > \frac{3 + 2 \sqrt{6} + 2}{3 - 2} \\ \\ = > \frac{5 + 2 \sqrt{6} }{1} \\ \\ = > 5 + 2 \sqrt{6}$

3)

$\frac{3}{5 - \sqrt{3} } + \frac{2}{5 + \sqrt{ 3} } \\ raionalizing \: the \: denominator - \\ \\ \frac{3}{ 5 - \sqrt{3} } \times \frac{5 + \sqrt{3} }{5 + \sqrt{3} } \\ \\ = > \frac{15 + 3}{25 - 3} \\ \\ = > \frac{18}{22} \\ \\ \\ \frac{2}{5 + \sqrt{3} } \times \frac{5 - \sqrt{3} }{5 - \sqrt{ 3} } \\ \\ = > \frac{10 - 2 \sqrt{3} }{25 - 3} \\ \\ = > \frac{10 - 2 \sqrt{3} }{22} \\ therefore \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \frac{3}{5 - \sqrt{3} } + \frac{2}{5 + \sqrt{3} } \\ \\ = > \frac{18}{22} + \frac{10 - 2 \sqrt{3} }{22} \\ \\ = > \frac{18 + 10 - 2 \sqrt{3} }{22} \\ \\ = > \frac{28 - 2 \sqrt{3} }{22} \\ \\ = > \frac{28}{22} - \frac{2 \sqrt{3} }{22} \\ \\ = > \frac{14}{11} - \frac{ \sqrt{3} }{11} \\ \\ = > \frac{14 - \sqrt{3} }{11}$