Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]\frac{1+\sqrt{2} }{3-\sqrt{2} } =\frac{(1+\sqrt{2} )(3+\sqrt{2} )}{(3-\sqrt{2})(3+\sqrt{2}) }=\frac{3+\sqrt{2}+3\sqrt{2} +2 }{9-2} =\frac{5+4\sqrt{2} }{7} =\frac{5}{7} +\frac{4}{7} \sqrt{2}[/tex]
Odpowiedź:
[tex]\dfrac{1+\sqrt{2}}{3-\sqrt{2}}=\dfrac{1+\sqrt{2}}{3-\sqrt{2}}\cdot\dfrac{3+\sqrt{2}}{3+\sqrt{2}}=\dfrac{(1+\sqrt{2})(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}=\dfrac{3+\sqrt{2}+3\sqrt{2}+2}{3^2-(\sqrt{2})^2}=\\\\\\=\dfrac{5+4\sqrt{2}}{9-2}=\dfrac{5+4\sqrt{2}}{7}[/tex]
