Question

answer this question plz hurry

Answer 1

First of all rationalising

$\frac{2 \sqrt{5} + \sqrt{3} }{2 \sqrt{5} - 3 } \\ \\ = > \frac{2 \sqrt{5} + \sqrt{3} }{2 \sqrt{5} - \sqrt{3} } \times \frac{2 \sqrt{5} + \sqrt{3} }{2 \sqrt{5} + \sqrt{3} } \\ \\ = > \frac{(2 \sqrt{5} + \sqrt{3} ) {}^{2} }{( 2\sqrt{5}) {}^{2} - ( \sqrt{3}) {}^{2} } \\ \\ = > \frac{(2 \sqrt{5}) {}^{2} + ( \sqrt{3}) {}^{2} + 2(2 \sqrt{5} )( \sqrt{3} )}{20 - 3} \\ \\ = > \frac{20 + 3 + 4 \sqrt{15} }{17}$
Similarly,

$\frac{2 \sqrt{5} - \sqrt{3} }{2 \sqrt{5} + \sqrt{3} } \\ \\ = \frac{2 \sqrt{5} - \sqrt{3} }{2 \sqrt{5} + \sqrt{3} } \times \frac{2 \sqrt{5} - \sqrt{3} }{2 \sqrt{5} - \sqrt{3} } \\ \\ = \frac{(2 \sqrt{5} - \sqrt{3} ) {}^{2} }{(2 \sqrt{5}) {}^{2} - ( \sqrt{3}) {}^{2} } \\ \\ = > \frac{20 + 3 - 4 \sqrt{15} }{17}$
adding them, we get

$\frac{20 + 3 + 4 \sqrt{15} }{17} + \frac{20 + 3 - 4 \sqrt{15} }{17}$
=
$\frac{23 + 23 + 4 \sqrt{15} - 4 \sqrt{15} }{17} \\ = > \frac{46}{17} = a + b \sqrt{15}$
comparing, we get

$a = \frac{46}{17} \\ b = 0$
Hope it helps dear friend ☺️

Answer 2

here is it !
hope it helps !