$$\begin{gathered} {\displaystyle \frac{(2+\sqrt{2})(3-\sqrt{5})}{(3+\sqrt{5})(2-\sqrt{2})}=\frac{{(2+\sqrt{2})}^2{(3-\sqrt{5})}^2}{(9-5)(4-2)}=\frac{(4+2+4\sqrt{2})(9+5-6\sqrt{5})}{8} \\ <div><br\mathrm{/}><\mathrm{/}div><div>} \end{gathered}$$\displaystyle =\frac { (6+4\sqrt { 2 } )(14-6\sqrt { 5 } ) }{ 8 } =\frac { (3+2\sqrt { 2 } )(7-3\sqrt { 5 } ) }{ 2 } =\frac { (21-9\sqrt { 5 } +14\sqrt { 2 } -6\sqrt { 10 } ) }{ 2 } \\ The\quad above\quad given\quad expression\quad consists\quad of\quad an\quad algebraic\quad equation\quad in\quad numerator\quad \\ consisting\quad of\quad irrational\quad terms,\quad hence\quad it\quad is\quad an\quad irrational\quad expression.\\ $$
The given statement is false.