[tex]a) \ \frac{1}{\sqrt{2}}=\frac{1\cdot\sqrt{2}}{\sqrt{2}\cdot\sqrt{2}} = \frac{\sqrt{2}}{2}[/tex]
[tex]b) \ \frac{1}{2-\sqrt{3}}=\frac{1\cdot(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})} =\frac{2+\sqrt{3}}{4-3} = \frac{2+\sqrt{3}}{1} = 2+\sqrt{3}[/tex]
[tex]c) \ \frac{4}{2-\sqrt{2}}=\frac{4(2+\sqrt{2})}{(2-\sqrt{2})(2+\sqrt{2})} =\frac{4(2+\sqrt{2})}{4-2} =\frac{4(2+\sqrt{2})}{2} = 2(2+\sqrt{2}) = 4+2\sqrt{2}[/tex]
[tex]d) \ \frac{\sqrt{2}}{\sqrt{3}+2} =\frac{\sqrt{2}(\sqrt{3}-2)}{(\sqrt{3}+2)(\sqrt{3}-2)} =\frac{\sqrt{2\cdot3}-2\sqrt{2}}{3-4} =\frac{\sqrt{6}-2\sqrt{2}}{-1} = 2\sqrt{2}-\sqrt{6}[/tex]
[tex]e) \ \frac{3+2\sqrt{3}}{2\sqrt{3}-3}=\frac{(2\sqrt{3}+3)(2\sqrt{3}+\sqrt{3})}{(2\sqrt{3}-3)(2\sqrt{3}+3)} = \frac{(2\sqrt{3}+3)^{2}}{12-9} = \frac{12+12\sqrt{3}+9}{3} = \frac{21+12\sqrt{3}}{3} = 7+4\sqrt{3}[/tex]
Wykorzystano wzory skróconego mnożenia:
[tex](a+b)(a-b) = a^{2}-b^{2}\\\\(a+b)^{2} = a^{2}+2ab + b^{2}[/tex]
