1) Oblicz: [latex](1+ 2sqrt{3})^2 - ( sqrt{3}-2)^2+(1- sqrt{7} )(1+ sqrt{7}) [/latex] Wynik: [latex] 8sqrt{3} [/latex] 2) Oblicz(skrócone mnożenie): [latex]( sqrt{3-2 sqrt{2} } - sqrt{3+2 sqrt{2} } )^2[/latex] Wynik: 4

[latex](1+2sqrt{3})^{2}-(sqrt{3}-2)^{2}+(1-sqrt{7})(1+sqrt{7})=\ \ =1+2*1*2sqrt{3}+(2sqrt{3})^{2}-(sqrt{3}^{2}-2*sqrt{3}*2+2^{2})+1-sqrt{7}^{2}=\ \ =1+4sqrt{3}+2^{2}*sqrt{3}^{2}-(3-4sqrt{3}+4)+1-7=\ \ =1+4sqrt{3}+4*3-(7-4sqrt{3})+1-7=\ \ =1+4sqrt{3}+12-7+4sqrt{3}+1-7=\ \ =8sqrt{3}[/latex] [latex]Wzory:\ (a+b)^{2}=a^{2}+2ab+b^{2}\ (a-b)^{2}=a^{2}-2ab+b^{2}\ (a-b)(a+b)=a^{2}-b^{2}[/latex] ======================================================== [latex](sqrt{3-2sqrt{2}}-sqrt{3+2sqrt{2}})^{2}=\ \ =(sqrt{1-2*1*sqrt{2}+sqrt{2}^{2}}-sqrt{1+2*1*sqrt{2}+sqrt{2}^{2}})^{2}=\ \ =(sqrt{(1-sqrt{2})^{2}}-sqrt{(1+sqrt{2})^{2}})^{2}=\ \ =(|1-sqrt{2}|-|1+sqrt{2}|)^{2}=\[/latex] ---> Wartość bezwzględna zawsze dodatnia: √2≈1,41 |1-√2|=√2-1 [latex]=(sqrt{2}-1-(1+sqrt{2}))^{2}=\ \ =(sqrt{2}-1-1-sqrt{2})^{2}=\ \ =(-2)^{2}=\ \ =4[/latex]