If 2 root 3 + 3 root 2 / 2 root 3 - 3 root = a+b root 6 , then find th

1.	If 2 root 3 + 3 root 2 / 2 root 3 - 3 root = a+b root 6 , then find the value of a and b plz.......answer this fast...........i have exam tmrow....plz
Answer» We've been GIVEN that; $\Longrightarrow \sf \dfrac{2\sqrt{3} + 3\sqrt{2}}{2\sqrt{3} - 3\sqrt{2}} = a + <klux>B</klux>\sqrt{6}$ LHS; $\Longrightarrow \sf \dfrac{2\sqrt{3} + 3\sqrt{2}}{2\sqrt{3} - 3\sqrt{2}}$ $\Longrightarrow \sf \dfrac{2\sqrt{3} + 3\sqrt{2}}{2\sqrt{3} - 3\sqrt{2}} \times \dfrac{2\sqrt{3} + 3\sqrt{2}}{2\sqrt{3} + 3\sqrt{2}}$ $\Longrightarrow \sf \dfrac{\Big(2\sqrt{3} + 3\sqrt{2}\Big)^2}{\Big(2\sqrt{3}\Big)^2 - \Big(3\sqrt{2}\Big)^2}$ $\Longrightarrow \sf \dfrac{4(3) + 9(2) + 12\sqrt{6}}{4(3) - 9(2)}$ $\Longrightarrow \sf \dfrac{12 + 18 + 12\sqrt{6}}{12 - 18}$ $\Longrightarrow \sf \dfrac{30 + 12\sqrt{6}}{-6}$ $\Longrightarrow \sf \dfrac{6(5 + 2\sqrt{6})}{-6}$ $\Longrightarrow \sf \dfrac{5 + 2\sqrt{6}}{-<klux>1</klux>}$ $\Longrightarrow \sf \dfrac{5}{-1} + \dfrac{2\sqrt{6}}{-1}$ ⇒ a = 5/-1 ⇒ b = (2√6)/-1