Oblicz: 1) [latex] sqrt{13-4 sqrt{3} } - sqrt{28+6 sqrt{3} } =[/latex] 2) [latex] sqrt{11-4 sqrt{7} } - sqrt{29-4 sqrt{7} } =[/latex] jest to zad 3.89 ze str. 84 c) i d) podr pazdro, matma 1 liceum

(a+b)² = a² + 2ab + b²  (a-b)² = a² - 2ab + b²  √a²=|a| |x|=x dla x≥0 |x|=-x  x<0 1) [latex] sqrt{13-4 sqrt{3}}- sqrt{28+6 sqrt{3}}= \ \ = sqrt{12-2*2 sqrt{3}+1}- sqrt{27+2*3sqrt{3}+1}= \ \ = sqrt{(2 sqrt{3})^{2}-2*2 sqrt{3}*1+1^{2} } -sqrt{(3sqrt{3})^{2}+2*3sqrt{3}*1+1}= \ \ = sqrt{(2sqrt{3}-1)^{2}} - sqrt{(3sqrt{3}+1)^{2}} =|2sqrt{3}-1|-|3sqrt{3}+1|=[/latex] [latex]=2sqrt{3}-1-(3sqrt{3}+1)=2sqrt{3}-1-3sqrt{3}-1=-sqrt{3}-2[/latex] 2) [latex] sqrt{11-4 sqrt{7}}- sqrt{29-4 sqrt{7}}= \ \ = sqrt{7-2*2*sqrt{7}+4}- sqrt{28+2*2sqrt{7}*1+1}= \ \ = sqrt{(sqrt{7})^{2}-2*2*sqrt{7}+2^{2} } -sqrt{(2sqrt{7})^{2}-2*2sqrt{7}*1+1^{2}}= \ \ = sqrt{(sqrt{7}-2)^{2}} - sqrt{(2sqrt{7}-1)^{2}} =|sqrt{7}-2|-|2sqrt{7}-1|=[/latex] [latex]=sqrt{7}-1-(2sqrt{7}-1)=sqrt{7}-1-2sqrt{7}+1=-sqrt{7}[/latex]