Question

Proszę o pomoc, usuń niewymierność z mianownika

Answer 1

Odpowiedź:

[tex]a)\ \ \dfrac{1}{6+\sqrt{2}}=\dfrac{1}{6+\sqrt{2}}\cdot\dfrac{6-\sqrt{2}}{6-\sqrt{2}}=\dfrac{6-\sqrt{2}}{(6+\sqrt{2})(6-\sqrt{2})}=\dfrac{6-\sqrt{2}}{6^2-(\sqrt{2})^2}=\dfrac{6-\sqrt{2}}{36-2}=\\\\\\=\dfrac{6-\sqrt{2}}{34}[/tex]

[tex]b)\ \ \dfrac{2\sqrt{2}}{\sqrt{5}+4}=\dfrac{2\sqrt{2}}{\sqrt{5}+4}\cdot\dfrac{\sqrt{5}-4}{\sqrt{5}-4}=\dfrac{2\sqrt{2}(\sqrt{5}-4)}{(\sqrt{5}+4)(\sqrt{5}-4)}=\dfrac{2\sqrt{10}-8\sqrt{2}}{(\sqrt{5})^2-4^2}=\\\\\\=\dfrac{2\sqrt{10}-8\sqrt{2}}{5-16}=\dfrac{2\sqrt{10}-8\sqrt{2}}{-11}=-\dfrac{2\sqrt{10}-8\sqrt{2}}{11}[/tex]

[tex]c)\ \ \dfrac{2}{4-3\sqrt{2}}=\dfrac{2}{4-3\sqrt{2}}\cdot\dfrac{4+3\sqrt{2}}{4+3\sqrt{2}}=\dfrac{2(4+3\sqrt{2})}{(4-3\sqrt{2})(4+3\sqrt{2})}=\dfrac{2(4+3\sqrt{2})}{4^2-(3\sqrt{2})^2}=\\\\\\=\dfrac{2(4+3\sqrt{2})}{16-9\cdot2}=\dfrac{2(4+3\sqrt{2})}{16-18}=\dfrac{\not2(4+3\sqrt{2})}{-\not2}=-(4+3\sqrt{2})=-4-3\sqrt{2}[/tex]

[tex]d)\ \ \dfrac{2\sqrt{3}-1}{2\sqrt{3}+2}=\dfrac{2\sqrt{3}-1}{2\sqrt{3}+2}\cdot\dfrac{2\sqrt{3}-2}{2\sqrt{3}-2}=\dfrac{(2\sqrt{3}-1)(2\sqrt{3}-2)}{(2\sqrt{3}+2)(2\sqrt{3}-2)}=\dfrac{4\cdot3-4\sqrt{3}-2\sqrt{3}+2}{(2\sqrt{3})^2-2^2}=\\\\=\dfrac{12-6\sqrt{3}+2}{4\cdot3-4}=\dfrac{14-6\sqrt{3}}{12-4}=\dfrac{14-6\sqrt{3}}{8}=\dfrac{\not2^1(7-3\sqrt{3})}{\not8_{4}}=\dfrac{7-3\sqrt{3}}{4}[/tex]

[tex]e)\ \ \dfrac{\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\dfrac{\sqrt{3}}{\sqrt{5}+\sqrt{3}}\cdot\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\dfrac{\sqrt{3}(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}=\dfrac{\sqrt{15}-3}{(\sqrt{5})^2-(\sqrt{3})^2}=\\\\=\dfrac{\sqrt{15}-3}{5-3}=\dfrac{\sqrt{15}-3}{2}[/tex]

Aby usunąć niewymierność mnożymy licznik i mianownik przez wyrażenie znajdujące się w mianowniku i zmieniamy znak wewnątrz wyrażenia, aby zastosować wzór skróconego mnożenia.

[tex](a-b)(a+b)=a^2-b^2[/tex]