Witam, jedno zadanko w załączniku. Dam NAJ! :)

[latex]x+y=4-2sqrt{3}+3+sqrt{3}=7-sqrt{3}\ \ x-y=4-2sqrt{3}-(3+sqrt{3})=4-2sqrt{3}-3-sqrt{3}=1-3sqrt{3}\ \ x cdot y=(4-2sqrt{3})(3+sqrt{3})=12+4sqrt{3}-6sqrt{3}-6=6-2sqrt{3}\ \ frac{x}{y}=frac{4-2sqrt{3}}{3+sqrt{3}}=frac{(4-2sqrt{3})(3-sqrt{3})}{(3+sqrt{3})(3-sqrt{3})}=\ =frac{12-4sqrt{3}-6sqrt{3}+6}{9-3}=frac{18-10sqrt{3}}{6}=3-frac{5sqrt{3}}{3}[/latex]

Suma: [latex]4 - 2sqrt{3} + 3 + sqrt{3} = 7 - sqrt{3} [/latex] Różnica: [latex]4 - 2sqrt{3} - (3 + sqrt{3}) = 4 - 2sqrt{3} - 3 - sqrt{3} = 1 - 3 sqrt{3} [/latex] Iloczyn: [latex](4 - 2sqrt{3}) * (3 + sqrt{3}) = 12 + 4 sqrt{3} - 6 sqrt{3} - 6 = 6 - 2 sqrt{3} [/latex] Iloraz: [latex](4 - 2sqrt{3}) : (3 + sqrt{3}) = frac{4}{3} + frac{4 sqrt{3} }{3} - frac{2 sqrt{3} }{3} - 2 = frac{4+2 sqrt{3} }{3} - 2 [/latex]