Pokażemy, że po wykonaniu obliczeń (dodawanie lub odejmowanie liczb) oraz usunięciu niewymierności

z mianowników ułamków otrzymane wyniki będą liczbami całkowitymi.

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$\text{a)}\frac{1}{\sqrt{2}-1}-\sqrt{2}=\frac{1}{\sqrt{2}-1}\cdot \frac{\sqrt{2}+1}{\sqrt{2}+1}-\sqrt{2}=\frac{\sqrt{2}+1}{2-1}-\sqrt{2}=$  

$=\sqrt{2}+1-\sqrt{2}=1$  

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$\text{b)}\frac{\sqrt{3}-2}{\sqrt{3}+2}-\frac{12}{\sqrt{3}}=\frac{\sqrt{3}-2}{\sqrt{3}+2}\cdot \frac{\sqrt{3}-2}{\sqrt{3}-2}-\frac{12}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=$   

$=\frac{{\left(\sqrt{3}-2\right)}^2}{3-4}-\frac{12\sqrt{3}}{3}=\frac{3-4\sqrt{3}+4}{-1}-4\sqrt{3}=\frac{7-4\sqrt{3}}{-1}-4\sqrt{3}=$  

$=-7+4\sqrt{3}-4\sqrt{3}=-7$  

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$\text{c)}\frac{2\sqrt{6}-1}{\sqrt{6}+3}-\sqrt{32\frac{2}{3}}=\frac{2\sqrt{6}-1}{\sqrt{6}+3}\cdot \frac{\sqrt{6}-3}{\sqrt{6}-3}-\sqrt{\frac{98}{3}}=$  

$=\frac{\left(2\sqrt{6}-1\right)\left(\sqrt{6}-3\right)}{6-9}-\frac{\sqrt{98}}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{12-6\sqrt{6}-\sqrt{6}+3}{-3}-\frac{\sqrt{294}}{3}=$  

$=\frac{15-7\sqrt{6}}{-3}-\frac{7\sqrt{6}}{3}=\frac{7\sqrt{6}-15}{3}-\frac{7\sqrt{6}}{3}=\frac{7\sqrt{6}-15-7\sqrt{6}}{3}=-\frac{15}{3}=-5$  

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$\text{d)}\frac{1}{3+\sqrt{6}}+\frac{1}{3-\sqrt{6}}=\frac{1}{3+\sqrt{6}}\cdot \frac{3-\sqrt{6}}{3-\sqrt{6}}+\frac{1}{3-\sqrt{6}}\cdot \frac{3+\sqrt{6}}{3+\sqrt{6}}=$  

$=\frac{3-\sqrt{6}}{9-6}+\frac{3+\sqrt{6}}{9-6}=\frac{3-\sqrt{6}}{3}+\frac{3+\sqrt{6}}{3}=\frac{3-\sqrt{6}+3+\sqrt{6}}{3}=\frac{6}{3}=2$  

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$\text{e)}\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}=\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}\cdot \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\cdot \frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}=$  

$=\frac{{\left(\sqrt{7}+\sqrt{5}\right)}^2}{7-5}+\frac{{\left(\sqrt{7}-\sqrt{5}\right)}^2}{7-5}=\frac{7+2\sqrt{35}+5}{2}+\frac{7-2\sqrt{35}+5}{2}=$  

$=\frac{12+2\sqrt{35}+12-2\sqrt{35}}{2}=\frac{24}{2}=12$  

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$\text{f)}\frac{\sqrt{6}-\sqrt{3}}{\sqrt{6}+\sqrt{3}}-\frac{\sqrt{3}+\sqrt{6}}{\sqrt{3}-\sqrt{6}}=\frac{\sqrt{6}-\sqrt{3}}{\sqrt{6}+\sqrt{3}}\cdot \frac{\sqrt{6}-\sqrt{3}}{\sqrt{6}-\sqrt{3}}-\frac{\sqrt{3}+\sqrt{6}}{\sqrt{3}-\sqrt{6}}\cdot \frac{\sqrt{3}+\sqrt{6}}{\sqrt{3}+\sqrt{6}}=$  

$=\frac{{\left(\sqrt{6}-\sqrt{3}\right)}^2}{6-3}-\frac{{\left(\sqrt{3}+\sqrt{6}\right)}^2}{3-6}=\frac{6-2\sqrt{18}+3}{3}-\frac{3+2\sqrt{18}+6}{-3}=$  

$=\frac{9-2\sqrt{18}}{3}+\frac{9+2\sqrt{18}}{3}=\frac{\left(9-2\sqrt{18}\right)+9+2\sqrt{18}}{3}=\frac{18}{3}=6$