Question

please send solution also​

Answer 1

Answer:

Given:

$\dfrac{\sqrt3 - 1}{\sqrt3+1} = a + b\sqrt3$

Solution:

$\implies \dfrac{\sqrt3 - 1}{\sqrt3 + 1} = a + b\sqrt3$

$\implies \dfrac{\sqrt3-1}{\sqrt3 + 1} \times \dfrac{\sqrt3 - 1}{ \sqrt3 - 1} = a + b\sqrt3$

$\implies \dfrac{(\sqrt3 - 1)^2}{(\sqrt3)^2 - 1^2} = a + b \sqrt3$

$\implies \dfrac{3 + 1 - 2\sqrt3}{3 - 1} = a + b \sqrt3$

$\implies \dfrac{4-2\sqrt3}{2} = a + b \sqrt3$

$\implies 2 - \sqrt3 = a+b \sqrt3$

$\boxed{\implies{\boxed{a = 2 , \ b = - 1}}}$