Odpowiedź:
[tex]1)\ \ (3+8\sqrt{3})^2=3^2+2\cdot3\cdot8\sqrt{3}+(8\sqrt{3})^2=9+48\sqrt{3}+64\cdot3=9+48\sqrt{3}+192=\\\\=201+48\sqrt{3}\\\\\\2)\ \ (3-2\sqrt{5})^2=3^2-2\cdot3\cdot2\sqrt{5}+(2\sqrt{5})^2=9-12\sqrt{5}+4\cdot5=9-12\sqrt{5}+20=\\\\=29-12\sqrt{5}\\\\\\3)\ \ (1-6\sqrt{7})^2=1^2-2\cdot1\cdot6\sqrt{7}+(6\sqrt{7})^2=1-12\sqrt{7}+36\cdot7=1-12\sqrt{7}+252=\\\\=253-12\sqrt{7}\\\\\\4)\ \ (\sqrt{2}+3)^2=(\sqrt{2})^2+2\sqrt{2}\cdot3+3^2=2+6\sqrt{2}+9=11+6\sqrt{2}[/tex]
[tex]5)\ \ (6-\sqrt{11})^2=6^2-2\cdot6\sqrt{11}+(\sqrt{11})^2=36-12\sqrt{11}+11=47-12\sqrt{11}\\\\\\6)\ \ (2\sqrt{3}-\sqrt{5})^2=(2\sqrt{3})^2-2\cdot2\sqrt{3}\cdot\sqrt{5}+(\sqrt{5})^2=4\cdot3-4\sqrt{15}+5=\\\\=12-4\sqrt{15}+5=17-4\sqrt{15}\\\\\\7)\ \ (9-\sqrt{6})^2=9^2-2\cdot9\cdot\sqrt{6}+(\sqrt{6})^2=81-18\sqrt{6}+6=87-18\sqrt{6}[/tex]
[tex]Zastosowane\ \ wzory\\\\(a+b)^2=a^2+2ab+b^2\\\\(a-b)=a^2-2ab+b^2[/tex]
