Question

Please find the answer , it is a very importatnt question

Answer 1

Step-by-step explanation:

The given expression is :

$\dfrac{2-\sqrt5}{2+3\sqrt5}=\sqrt5 x+y$ ....(1)

Taking LHS of the above expression

$\dfrac{2-\sqrt5}{2+3\sqrt5}$

Rationalizing both denominator and numerator.

$\dfrac{2-\sqrt5}{2+3\sqrt5}\times \dfrac{2-3\sqrt5}{2-3\sqrt5}=\dfrac{4-6\sqrt5-2\sqrt5+15}{2^2-(3\sqrt5)^2}\\\\=\dfrac{19-8\sqrt5}{4-45}\\\\=\dfrac{19-8\sqrt5}{-41}\\\\=\dfrac{\sqrt5 \times 8}{41}+(\dfrac{-19}{41})$

Equation (1) becomes,

$\dfrac{\sqrt5 \times 8}{41}+(\dfrac{-19}{41})=\sqrt 5\ x+y$

Equating both sides,

$x=\dfrac{8}{41}$ and $y=\dfrac{-19}{41}$

Hence, this is the required solution.