W tym zadaniu skorzystamy ze wzoru na:

• kwadrat sumy:
  
  ${\left(a+b\right)}^2=a^2+2ab+b^2$
  
  
• kwadrat różnicy:
  
  ${\left(a-b\right)}^2=a^2-2ab+b^2$
  
  
• różnicę kwadratów:
  
  $a^2-b^2=\left(a-b\right)\left(a+b\right)$

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$\text{a)}{\left(1+\sqrt{2}\right)}^2+{\left(1-\sqrt{2}\right)}^2=\left(1+2\sqrt{2}+2\right)+\left(1-2\sqrt{2}+2\right)=$

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

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$\text{b)}{\left(\sqrt{3}-1\right)}^2-{\left(2-\sqrt{3}\right)}^2=\left(3-2\sqrt{3}+1\right)-\left(4-4\sqrt{3}+3\right)=$

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

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$\text{c)}\left(4-\sqrt{5}\right)\left(4+\sqrt{5}\right)-\left(\sqrt{5}-2\right)\left(2+\sqrt{5}\right)=$

$=4^2-{\sqrt{5}}^2-\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=$   

$=16-5-\left({\sqrt{5}}^2-2^2\right)=11-\left(5-4\right)=11-1=10$

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$\text{d)}\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)+{\left(\sqrt{6}-\sqrt{5}\right)}^2=$

$=\left({\sqrt{6}}^2-{\sqrt{5}}^2\right)+\left(6-2\sqrt{6}\cdot \sqrt{5}+5\right)=$   

$=\left(6-5\right)+\left(11-2\sqrt{30}\right)=1+11-2\sqrt{30}=12-2\sqrt{30}$