Step-by-step explanation:
The given expression is :
[tex]\dfrac{2-\sqrt5}{2+3\sqrt5}=\sqrt5 x+y[/tex] ....(1)
Taking LHS of the above expression
[tex]\dfrac{2-\sqrt5}{2+3\sqrt5}[/tex]
Rationalizing both denominator and numerator.
[tex]\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})[/tex]
Equation (1) becomes,
[tex]\dfrac{\sqrt5 \times 8}{41}+(\dfrac{-19}{41})=\sqrt 5\ x+y[/tex]
Equating both sides,
[tex]x=\dfrac{8}{41}[/tex] and [tex]y=\dfrac{-19}{41}[/tex]
Hence, this is the required solution.
