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[tex]\sqrt{a}\cdot\sqrt{a}=a\ \text{dla}\ a\geq0\\\\(\sqrt{a})^2=a\ \text{dla}\ a\geq0\\\\(a-b)(a+b)=a^2-b^2[/tex]
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[tex]a)\ \dfrac{3-\sqrt3}{\sqrt3}=\dfrac{3-\sqrt3}{\sqrt3}\cdot\dfrac{\sqrt3}{\sqrt3}=\dfrac{3\sqrt3-3}{3}=\dfrac{3(\sqrt3-1)}{3}=\sqrt3-1\\\\b)\ \dfrac{2+\sqrt2}{\sqrt2}=\dfrac{2+\sqrt2}{\sqrt2}\cdot\dfrac{\sqrt2}{\sqrt2}=\dfrac{2\sqrt2+2}{2}=\dfrac{2(\sqrt2+1)}{2}=\sqrt2+1\\\\c)\ \dfrac{2-\sqrt[3]4}{\sqrt[3]4}=\dfrac{2-\sqrt[3]4}{\sqrt[3]{2^2}}\cdot\dfrac{\sqrt[3]2}{\sqrt[3]2}=\dfrac{2\sqrt[3]2-\sqrt[3]8}{\sqrt[3]{2^3}}=\dfrac{2\sqrt[3]2-2}{2}=\dfrac{2(\sqrt[3]2-1)}{2}=\sqrt[3]2-1[/tex]
[tex]d)\ \dfrac{6+\sqrt[3]9}{\sqrt[3]9}=\dfrac{6+\sqrt[3]9}{\sqrt[3]{3^2}}\cdot\dfrac{\sqrt[3]3}{\sqrt[3]3}=\dfrac{6\sqrt[3]3+\sqrt[3]{27}}{3}=\dfrac{6\sqrt[3]3+3}{3}=\dfrac{3(2\sqrt[3]3+1)}{3}=2\sqrt[3]3+1\\\\e)\ \dfrac{\sqrt6+2}{\sqrt6-2}=\dfrac{\sqrt6+2}{\sqrt6-2}\cdot\dfrac{\sqrt6+2}{\sqrt6+2}=\dfrac{(\sqrt6+2)^2}{(\sqrt6)^2-2^2}=\dfrac{(\sqrt6)^2+2\cdot\sqrt6\cdot2+2^2}{6-4}\\\\=\dfrac{6+4\sqrt6+4}{2}=\dfrac{10+4\sqrt6}{2}=\dfrac{2(5+2\sqrt6)}{2}=5+2\sqrt6[/tex]
[tex]f)\ \dfrac{\sqrt5-\sqrt2}{\sqrt5+\sqrt2}=\dfrac{\sqrt5-\sqrt2}{\sqrt5+\sqrt2}\cdot\dfrac{\sqrt5-\sqrt2}{\sqrt5-\sqrt2}=\dfrac{(\sqrt5-\sqrt2)^2}{(\sqrt5)^2-(\sqrt2)^2}=\dfrac{(\sqrt5)^2-2\cdot\sqrt5\cdot\sqrt2+(\sqrt2)^2}{5-2}\\\\=\dfrac{5-2\sqrt{10}+2}{3}=\dfrac{7-2\sqrt{10}}{3}\\\\g)\ \dfrac{\sqrt3+2}{\sqrt3-2}=\dfrac{\sqrt3+2}{\sqrt3-2}\cdot\dfrac{\sqrt3+2}{\sqrt3+2}=\dfrac{(\sqrt3+2)^2}{(\sqrt3)^2-2^2}=\dfrac{(\sqrt3)^2+2\cdot\sqrt3\cdot2+2^2}{3-4}\\\\=\dfrac{3+4\sqrt3+4}{-1}=-7-4\sqrt3[/tex]
[tex]h)\ \dfrac{2}{2-\sqrt3}=\dfrac{2}{2-\sqrt3}\cdot\dfrac{2+\sqrt3}{2+\sqrt3}=\dfrac{4+2\sqrt3}{2^2-(\sqrt3)^2}=\dfrac{4+2\sqrt3}{4-3}=4+2\sqrt3\\\\i)\ \dfrac{1+2\sqrt5}{\sqrt5+3}=\dfrac{1+2\sqrt5}{\sqrt5+3}\cdot\dfrac{\sqrt5-3}{\sqrt5-3}=\dfrac{\sqrt5+2\cdot5-3-6\sqrt5}{(\sqrt5)^2-3^2}=\dfrac{10-3-5\sqrt6}{5-9}\\\\=\dfrac{7-5\sqrt6}{-4}=\dfrac{5\sqrt6-7}{4}[/tex]
[tex]j)\ \dfrac{22}{2\sqrt5-3}=\dfrac{22}{2\sqrt5-3}\cdot\dfrac{2\sqrt5+3}{2\sqrt5+3}=\dfrac{44\sqrt5+66}{(2\sqrt5)^2-3^2}=\dfrac{44\sqrt5+66}{4\cdot5-9}\\\\=\dfrac{44\sqrt5+66}{20-9}=\dfrac{44\sqrt5+66}{11}=\dfrac{11(4\sqrt5+6)}{11}=4\sqrt5+6[/tex]
