Odpowiedź:
[tex]a)\\\\\dfrac{(7^{\frac{1}{2}}-9^{\frac{1}{2}})(\sqrt{7}+3)}{(12^{\frac{1}{2}}-3^{\frac{1}{2}})^2}=\dfrac{(\sqrt{7}-\sqrt{9})(\sqrt{7} +3)}{(\sqrt{12}-\sqrt{3})^2}=\dfrac{(\sqrt{7}-3)(\sqrt{7}+3)}{(\sqrt{4\cdot3}-\sqrt{3})^2}=\dfrac{(\sqrt{7})^2-3^2}{(2\sqrt{3}-\sqrt{3})^2}=\\\\=\dfrac{7-9}{(\sqrt{3})^2}=\dfrac{-2}{3}=\dfrac{2}{3}[/tex]
[tex]b)\\\\\Big[{\dfrac{(3^{\frac{1}{4}}-2^{\frac{1}{4}})(3^{\frac{1}{4}}+(\sqrt{2})^{\frac{1}{2}})}{\sqrt{2}-\sqrt{3}}\Big]^2=\Big[\dfrac{(\sqrt[4]{3}-\sqrt[4]{2})(\sqrt[4]{3}+(2^{\frac{1}{2}})^{\frac{1}{2}}}{\sqrt{2}-\sqrt{3}}\Big]^2=\Big[\dfrac{(\sqrt[4]{3}-\sqrt[4]{2})(\sqrt[4]{3}+2^{\frac{1}{4}})}{\sqrt{2}-\sqrt{3}}\Big]^2=[/tex]
[tex]=\Big[\dfrac{(\sqrt[4]{3}-\sqrt[4]{2})(\sqrt[4]{3}+\sqrt[4]{2})}{\sqrt{2}-\sqrt{3}}\Big]^2=\Big[\dfrac{(\sqrt[4]{3})^2-(\sqrt[4]{2})^2}{\sqrt{2}-\sqrt{3}}\Big]^2=\Big[\dfrac{(3^{\frac{1}{4}})^2-(2^{\frac{1}{4}})^2}{\sqrt{2}-\sqrt{3}}\Big]^2\\\\\\=\Big[\dfrac{3^{\frac{1}{2}}-2^{\frac{1}{2}}}{\sqrt{2}-\sqrt{3}}\Big]^2=\Big[\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{2}-\sqrt{3}}\Big]^2=\Big[\dfrac{\sqrt{3}-\sqrt{2}}{-(-\sqrt{2}+\sqrt{3}}\Big]^2=(-1)^2=1[/tex]
[tex]Zastosowano\ \ wzory\\\\a^{\frac{m}{n}}=\sqrt[n]{a^m}\\\\(a-b)(a+b)=a^2-b^2[/tex]
