Jeśli $x=1+\sqrt{3}$ oraz $y=3-4\sqrt{4}$, to:
| (01) | $x+y=(1+\sqrt{3})+(3-4\sqrt{3})=4-3\sqrt{3}$ |
| (02) | $x-y=(1+\sqrt{3})-(3-4\sqrt{3})=-2+5\sqrt{3}$ |
| (03) | $x\cdot y=(1+\sqrt{3})\cdot(3-4\sqrt{3})=1\cdot3-1\cdot4\sqrt{3} +3\cdot\sqrt{3}-4\sqrt{3}\cdot\sqrt{3}= 3-4\sqrt{3} +3\sqrt{3}-4\cdot 3=-9-\sqrt{3}$ |
| (04) | $x^2=(1+\sqrt{3})^2=1^2+2\cdot 1\cdot \sqrt{3}+(\sqrt{3})^2=1+2\sqrt{3}+3=4+2\sqrt{3}$ |
| (05) | $y^2=(3-4\sqrt{3})^2=3^2-2\cdot 3\cdot 4\sqrt{3}+(4\sqrt{3})^2=9-24\sqrt{3}+48=57-24\sqrt{3}$ |
| (06) | $x^3=(1+\sqrt{3})^3=1^3+3\cdot 1^2\cdot \sqrt{3}+3\cdot 1\cdot (\sqrt{3})^2+(\sqrt{3})^3= 1+3\sqrt{3}+9+3\sqrt{3}=10+6\sqrt{3}$ |
| (07) | $y^3=(3-4\sqrt{4})^3=3^3-3\cdot 3^2\cdot 4\sqrt{3}+3\cdot 3\cdot (4\sqrt{3})^2-(4\sqrt{3})^3= 27-108\sqrt{3}+432-192\sqrt{3}=459-300\sqrt{3}$ |
| (08) | $\frac{1}{x-1}=\frac{1}{(1+\sqrt{3})-1}=\frac{1}{\sqrt{3}}=\frac{1\cdot \sqrt{3}}{\sqrt{3}\cdot \sqrt{3}}=\frac{\sqrt{3}}{3}=0+\frac{1}{3} \sqrt{3}$ |
| (09) | $\frac{1}{3-y}=\frac{1}{3-(3-4\sqrt{4})}=\frac{1}{4\sqrt{3}}=\frac{1\cdot \sqrt{3}}{4\sqrt{3}\cdot \sqrt{3}}= \frac{\sqrt{3}}{12}=0+\frac{1}{12} \sqrt{3}$ |
| (10) | $\frac{1}{x}=\frac{1}{(1+\sqrt{3})}=\frac{1\cdot (1-\sqrt{3})}{(1+\sqrt{3})\cdot (1-\sqrt{3})}= \frac{1-\sqrt{3}}{1^2-(\sqrt{3})^2}=\frac{1-\sqrt{3}}{1-3}=\frac{1-\sqrt{3}}{-2}=-\frac{1}{2}+\frac{1}{2}\sqrt{3}$ |
| (11) | $\frac{1}{y}=\frac{1}{(3-4\sqrt{3})}=\frac{1\cdot (3+4\sqrt{3})}{(3-4\sqrt{3})\cdot (3+4\sqrt{3})}= \frac{3+4\sqrt{3}}{3^2-(4\sqrt{3})^2}=\frac{3+4\sqrt{3}}{9-48}=\frac{3+4\sqrt{3}}{-39}=-\frac{3}{39}-\frac{4}{39}\sqrt{3}=-\frac{1}{13}-\frac{4}{39}\sqrt{3}$ |
| (12) | $\frac{x}{y}=\frac{1+\sqrt{3}}{(3-4\sqrt{3})}=\frac{(1+\sqrt{3})\cdot (3+4\sqrt{3})}{(3-4\sqrt{3})\cdot (3+4\sqrt{3})}= \frac{1\cdot3+1\cdot4\sqrt{3}+3\cdot\sqrt{3}+4\sqrt{3}\cdot\sqrt{3}}{3^2-(4\sqrt{3})^2}=\frac{15+7\sqrt{3}}{9-48}=\frac{15+7\sqrt{3}}{-39}=-\frac{15}{39}-\frac{7}{39}\sqrt{3}=-\frac{5}{13}-\frac{7}{39}\sqrt{3}$ |
| (13) | $\frac{y}{x}= \frac{3-4\sqrt{3}}{1+\sqrt{3}}=\frac{(3-4\sqrt{3})\cdot (1-\sqrt{3})}{(1+\sqrt{3})\cdot (1-\sqrt{3})}= \frac{3\cdot 1-3\sqrt{3}-4\sqrt{3} \cdot 1 +4\sqrt{3} \cdot \sqrt{3} }{1^2-(\sqrt{3})^2}=\frac{15-7\sqrt{3}}{1-3}= \frac{15-7\sqrt{3}}{-2}=-\frac{15}{2}+\frac{7}{2}\sqrt{3}$ |