Step-by-step explanation:
As per the provided information in the given question, we have to rationalise the denominator of :
[tex] \longmapsto \bf { \dfrac{\sqrt{3} + 1}{2\sqrt{2} - \sqrt{3}} } \\ [/tex]
In order to rationalise the denominator of any fraction, we multiply the rationalising factor of the denominator with both the numerator and denominator of fraction.
○ Rationalising factor is nothing but the conjugate of the denominator. Here, the denominator is in the form of (a – b). So, the rationalising factor will be (a + b). Henceforth, rationalising factor here will be (2√2 + √3).
[tex] \longrightarrow \sf{\quad { \dfrac{\sqrt{3} + 1}{2\sqrt{2} - \sqrt{3}} \times \dfrac{2\sqrt{2} + \sqrt{3}}{2\sqrt{2} + \sqrt{3}} }} \\ [/tex]
Multiplying (2√2 + √3) with both the numerator and the denominator of the fraction.
[tex] \longrightarrow \sf{\quad { \dfrac{(\sqrt{3} + 1)(2\sqrt{2} + \sqrt{3})}{(2\sqrt{2} -\sqrt{3})(2\sqrt{2} + \sqrt{3})} }} \\ [/tex]
Using the distributive property in the numerator to perform multiplication easily. And, in the denominator using the property,
[tex] \longrightarrow \sf{\quad { \dfrac{\sqrt{3}(2\sqrt{2} + \sqrt{3}) + 1(2\sqrt{2} + \sqrt{3})}{(2\sqrt{2})^2 -(\sqrt{3})^2} }} \\ [/tex]
Simplifying further by performing multiplication in the numerator and writing the squares of the terms in the denominator.
[tex] \longrightarrow \sf{\quad { \dfrac{2\sqrt{6} + 3 + 2\sqrt{2} + \sqrt{3} }{(4 \times 2) - 3} }} \\ [/tex]
Performing multiplication in the denominator.
[tex] \longrightarrow \sf{\quad { \dfrac{2\sqrt{6} + 3 + 2\sqrt{2} + \sqrt{3} }{8- 3} }} \\ [/tex]
Performing subtraction in the denominator.
[tex] \longrightarrow \quad \underline{\boxed { \dfrac{ \textbf{\textsf{2}}\sqrt{\textbf{\textsf{6}}} + \textbf{\textsf{3}} + \textbf{\textsf{2}}\sqrt{\textbf{\textsf{2}}} + \sqrt{\textbf{\textsf{3}}} }{\textbf{\textsf{5}}} }} \\ [/tex]
[tex] \underline { \sf { Hence, \; rationalised!! }}\\ [/tex]
