Odpowiedź:
[tex]a)\ \ (\sqrt{2}+3)^2+(\sqrt{2}-3)^2=(\sqrt{2})^2+2\sqrt{2}\cdot3+3^2+((\sqrt{2})^2-2\sqrt{2}\cdot3+3^2)=\\\\=2+6\sqrt{2}+9+(2-6\sqrt{2}+9)=11+6\sqrt{2}+2-6\sqrt{2}+9=11+2+9=22\\\\\\b)\ \ (2+\sqrt{3})^2-(2-\sqrt{3})^2=2^2+2\cdot2\sqrt{3}+(\sqrt{3})^2-(2^2-2\cdot2\sqrt{3}+(\sqrt{3})^2)=\\\\=4+4\sqrt{3}+3-(4-4\sqrt{3}+3)=7+4\sqrt{3}-4+4\sqrt{3}-3=4\sqrt{3}+4\sqrt{3}=8\sqrt{3} [/tex]
[tex]c)\ \ (2\sqrt{3}-1)^2-(\sqrt{3}+2)^2=(2\sqrt{3})^2-2\cdot2\sqrt{3}+1^2-((\sqrt{3})^2+2\sqrt{3}\cdot2+2^2)=\\\\=4\cdot3-4\sqrt{3}+1-(3+4\sqrt{3}+4)=12-4\sqrt{3}+1-(7+4\sqrt{3})=\\\\=13-4\sqrt{3}-7-4\sqrt{3}=6-8\sqrt{3} [/tex]
[tex]d)\ \ (2\sqrt{3}-1)^2+(2\sqrt{3}+1)^2=(2\sqrt{3})^2-2\cdot2\sqrt{3}+1^2+(2\sqrt{3})^2+2\cdot2\sqrt{3}+1^2=\\\\=4\cdot3-4\sqrt{3}+1+4\cdot3+4\sqrt{3}+1=12-4\sqrt{3}+1+12+4\sqrt{3}+1=12+1+12+1=26\\\\\\Zastosowane\ \ wzory\\\\(a+b)^2=a^2+2ab+b^2\\\\(a-b)=a^2-2ab+b^2 [/tex]
