[tex]x=3-\sqrt5\\y=4+2\sqrt5\\\\a)\\xy=(3-\sqrt5)(4+2\sqrt5)=12+6\sqrt5-4\sqrt5-10=2+2\sqrt5\\b) \\\frac{x}y=\frac{3-\sqrt5}{4+2\sqrt5}=\frac{3-\sqrt5}{4+2\sqrt5}*\frac{4-2\sqrt5}{4-2\sqrt5}=\frac{(3-\sqrt5)(4-2\sqrt5)}{4^2-(2\sqrt5)^2}=\frac{12-6\sqrt5-4\sqrt5+10}{16-20}=\frac{22-10\sqrt5}{-4}=-\frac{2(11-5\sqrt5)}4=-\frac{11-5\sqrt5}2[/tex]
[tex]c)\\3y-x^2=3(4+2\sqrt5)-(3-\sqrt5)^2=12+6\sqrt5-(9-6\sqrt5+5)=12+6\sqrt5-(14-6\sqrt5)=12+6\sqrt5-14+6\sqrt5=12\sqrt5-2[/tex]
[tex]d) \\x^2-y^2=(3-\sqrt5)^2-(4+2\sqrt5)^2=9-6\sqrt5+5-(16+16\sqrt5+20)=14-6\sqrt5-(36+16\sqrt5)=14-6\sqrt5-36-16\sqrt5=-22-22\sqrt5[/tex]
