Answer:
[tex]\frac{2\sqrt{3}+5\sqrt{2}}{3\sqrt{3}-2\sqrt{2} }[/tex] is equal to [tex]2+\sqrt{6}[/tex]
Step-by-step explanation:
Consider the given fraction,
[tex]\frac{2\sqrt{3}+5\sqrt{2}}{3\sqrt{3}-2\sqrt{2} }[/tex]
We are required to rational denominator,
Multiply and divide numerator by [tex]\Rightarrow \frac{2\sqrt{3}+5\sqrt{2}}{3\sqrt{3}-2\sqrt{2}} \times \frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}}[/tex] , we get,
Using identity ,[tex](a-b)(a+b)=a^2-b^2[/tex] in denominator, we get,
[tex]\Rightarrow \frac{2\sqrt{3}+5\sqrt{2}}{3\sqrt{3}-2\sqrt{2}} \times \frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}}=\frac{(2\sqrt{3}+5\sqrt{2})(3\sqrt{3}+2\sqrt{2})}{(3\sqrt{3})^2-(2\sqrt{2})^2)}[/tex]
[tex]\Rightarrow \frac{(2\sqrt{3}+5\sqrt{2})(3\sqrt{3}+2\sqrt{2})}{(3\sqrt{3})^2-(2\sqrt{2})^2)}=\frac{(2\sqrt{3}+5\sqrt{2})(3\sqrt{3}+2\sqrt{2})}{27-8}\\\\\\\\[/tex]
On solving further , we get,
[tex]\Rightarrow \frac{(2\sqrt{3}+5\sqrt{2})(3\sqrt{3}+2\sqrt{2})}{(3\sqrt{3})^2-(2\sqrt{2})^2)}=\frac{38+19\sqrt{6}}{19}[/tex]
solving , we get,
[tex]\Rightarrow \frac{38+19\sqrt{6}}{19}=2+\sqrt{6}[/tex]
Thus, [tex]\frac{2\sqrt{3}+5\sqrt{2}}{3\sqrt{3}-2\sqrt{2} }[/tex] is equal to [tex]2+\sqrt{6}[/tex]
