Question

find the value of A in each of the following 5+2√3/7+4√3=a-6√3

Answer 1

5+2√3/7+4√3=A-6√3
LHS
rationalise the LHS
5+2√3/7+4√3*7-4√3/7-4√3
by rationalisation we get
35-20√3+14√3-24/49-48
=11-6√3
A=11

Answer 2

Heya friend,
Here is the answer you were looking for:

Using the identity :

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

$\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } = a - 6 \sqrt{3}$

L.H.S

On rationalizing the denominator we get,

$= \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ \\ = \frac{5(7 - 4 \sqrt{3} ) + 2 \sqrt{3}(7 - 4 \sqrt{3} ) }{ {(7)}^{2} - {(4 \sqrt{3} )}^{2} } \\ \\ = \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 24 }{49 - 48} \\ \\ = 11 - 6 \sqrt{3}$

On comparing

$11 - 6 \sqrt{3} = a - 6 \sqrt{3} \\ \\ a = 11$

Hope this helps!!!

Feel free to ask in the comment section if you have any doubt regarding to my answer...

@Mahak24

Thanks...
☺☺