Odpowiedź:
[tex]a)\\(\sqrt{2}+1)^{2}-(2-\sqrt{2})^{2} = (\sqrt{2}^{2}+2\cdot\sqrt{2}\cdot1+1^{2})-(2^{2}-2\sqrt{2}+\sqrt{2}^{2})=\\\\=2+2\sqrt{2}+1-(4-2\sqrt{2}+2) = 3+2\sqrt{2}-6+4\sqrt{2} = 6\sqrt{2}-3[/tex]
[tex]b)\\(1-\sqrt{3})^{2}-(\sqrt{3}-2)^{2} = 1-2\sqrt{3}+3 -(3-4\sqrt{3}+4) = 4-2\sqrt{3}-7+4\sqrt{3}=\\\\=2\sqrt{3}-3[/tex]
[tex]c)\\(2+\sqrt{2})^{2}-(2\sqrt{2}-1)^{2} =4+4\sqrt{2}+2-(8-4\sqrt{2}+1) = 6+4\sqrt{2}-9+4\sqrt{2} =\\\\=8\sqrt{2}}-3[/tex]
[tex]d)\\(3-\sqrt{5})^{2}-(-1-\sqrt{5})^{2} =9-6\sqrt{5}+5-[-(1+\sqrt{5})]^{2} = 14-6\sqrt{5}-(1+2\sqrt{5}+5) =\\\\=14-6\sqrt{5}-(6+2\sqrt{5}) = 14-6\sqrt{5}-6-2\sqrt{5} = 8-8\sqrt{5}[/tex]
[tex]e)\\(4-\sqrt{5})^{2}-(-2+\sqrt{5})^{2} = 16-8\sqrt{5}+5-(4-4\sqrt{5}+5) =21-8\sqrt{5}-9+4\sqrt{5}=\\\\=12-4\sqrt{5}[/tex]
[tex]f)\\(1+\sqrt{6})^{2}-(-2-2\sqrt{6})^{2} = 1+2\sqrt{6}+6-[-(2+2\sqrt{6})]^{2} = 7+2\sqrt{6}-(4+8\sqrt{6}+24) =\\\\=7+2\sqrt{6}-28-8\sqrt{6}=-21-6\sqrt{6}[/tex]
[tex]g)\\(-\sqrt{2}-1)^{2}-(2\sqrt{2}+1)^{2} = [-(\sqrt{2}+1)]^{2}-(8+4\sqrt{2}+1) =2+2\sqrt{2}+1 -(9+4\sqrt{2}) =\\\\=3+2\sqrt{2}-9-4\sqrt{2} = -6-2\sqrt{2}[/tex]
[tex]h)\\(-3-\sqrt{3})^{2}-(1-2\sqrt{3})^{2} =[-(3+\sqrt{3})]^{2} -(1-4\sqrt{3}+12) =\\\\=9+6\sqrt{3}+3-(13-4\sqrt{3})=12+6\sqrt{3}-13+4\sqrt{3} = 10\sqrt{3}}-1[/tex]
[tex]i)\\(\sqrt{5}-1)^{2}-(-3-2\sqrt{5})^{2} = 5-2\sqrt{5}+1-[-(3+2\sqrt{5})]^{2} = 6-2\sqrt{5}-(9+12\sqrt{5}+20) =\\\\=6-2\sqrt{5}-29-12\sqrt{5} = -23-14\sqrt{5}[/tex]
Szczegółowe wyjaśnienie:
[tex](a+b)^{2} = a^{2}+2ab+b^{2}\\(a-b)^{2} = a^{2}-2ab + b^{2}[/tex]
