[tex]zad.3\\\\a=2+3\sqrt{5} ~~\land ~~b=4-\sqrt{5} \\\\a+b=2+3\sqrt{5} +4-\sqrt{5} =6+2\sqrt{5} =2\cdot (3+\sqrt{5} )\\\\a-b=2+3\sqrt{5} -(4-\sqrt{5})=2+3\sqrt{5} -4+\sqrt{5}=4\sqrt{5} -2=2\cdot ( 2\sqrt{5} -1) \\\\a\cdot b =(2+3\sqrt{5})\cdot (4-\sqrt{5}) =8-2\sqrt{5}+12\sqrt{5}-15=10\sqrt{5}-7\\\\\dfrac{a}{b} =\dfrac{2+3\sqrt{5}}{4-\sqrt{5}} \cdot \dfrac{4+\sqrt{5}}{4+\sqrt{5}}=\dfrac{8+2\sqrt{5}+12\sqrt{5}+15}{2^{2} -(\sqrt{5} )^{2} } =\dfrac{14\sqrt{5}+23 }{-1} =-14\sqrt{5}-23[/tex]
