Question

Please answer this question in deyail form
all the questions

Answer 1

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THE ANSWER IS HERE,


a)

$= > \: \frac{5 + 2\sqrt{3} }{7 + 4 \sqrt{3} }$

$= > \: \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} }$

$= > \: \frac{35 - 20 \sqrt{3 } + 14 \sqrt{3} - 24 }{ {7}^{2} - {(4 \sqrt{3})}^{2} }$

$= > \: \frac{11 - 6 \sqrt{3} }{49 - 48}$


$= > \: 11 - 6 \sqrt{3} .$


$comparing \: with \: \: a + b \sqrt{3}$

we get,

a=11 , b =-6


b)


$= > \: \frac{5 + \sqrt{3} }{ \sqrt{5} - \sqrt{3} }$

$= > \: \frac{5 + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } \times \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} + \sqrt{3} }$

$= > \: \frac{5 \sqrt{5} + 5 \sqrt{3} + \sqrt{15} + 3 }{5 - 3}$

$= > \: \frac{1}{2} (5 \sqrt{5} + 5 \sqrt{3} + 3) + \frac{1}{2} ( \sqrt{15} )$

$comparing \: with \: \: \frac{1}{2} a + 3b \sqrt{15}$

$a = 5 \sqrt{5} + 5 \sqrt{3} + 3$

$b = \frac{1}{6}$


19)

a)

$= > \: \frac{3}{ \sqrt{5} - \sqrt{3} }$

$= > \: \frac{3}{ \sqrt{5} - \sqrt{3} } \times \frac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} + \sqrt{3} }$

$= > \: \frac{3 \sqrt{5} + 3 \sqrt{3} }{5 - 3}$

$= > \: \frac{3 \sqrt{5} + 3 \sqrt{3} }{2}$

b)

$= > \: \frac{2 \sqrt{7} }{ \sqrt{5} + \sqrt{3} }$

$= > \: \frac{2 \sqrt{7} }{ \sqrt{5} + \sqrt{3} } \times \frac{ \sqrt{5} - \sqrt{3} }{ \sqrt{5} - \sqrt{3} }$

$= > \: \frac{2 \sqrt{35} - 2 \sqrt{21} }{5 - 3}$

$= > \: \frac{2 \sqrt{35} - 2 \sqrt{21} }{2}$

$= > \sqrt{35} - \sqrt{21}$


:-)Hope it helps u.

Answer 2

Answer:

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