Rozwiąz zad.13 i zad. 7 b)   (oblicz zad. 7) daje naj + 40 pkt

[latex]13.\a) (3-sqrt7)(2+2sqrt7)=6+6sqrt7-2sqrt7-2cdot7=oxed{4sqrt7-8}\\b) (sqrt3+2sqrt2)(4sqrt3-sqrt2)=4cdot3-sqrt6+8sqrt6-2cdot2=8+7sqrt6\\c) (sqrt6-2sqrt3)(2sqrt6+6sqrt3)=2cdot6+6sqrt{18}-4sqrt{18}-12cdot3\=2sqrt{18}-24=2sqrt{9cdot2}-24=2cdot3sqrt2-24=oxed{6sqrt2-24}\\d) (5sqrt2-2sqrt{10})(-3sqrt{10}-2sqrt2)=-15sqrt{20}-10cdot2+6cdot10+4sqrt{20}\=40-11sqrt{20}=40-11sqrt{4cdot5}=40-11cdot2sqrt5=oxed{40-22sqrt5}[/latex] [latex]7.\b)\sqrt[3]2cdotsqrt[3]4=sqrt[3]{8}=fbox2\\sqrt[3]{12}cdotsqrt[3]{18}=sqrt[3]{216}=fbox6\\sqrt[3]{24}cdotsqrt[3]{72}=sqrt[3]{8cdot3cdot8cdot9}=sqrt[3]{64cdot27}=4cdot3=fbox{12}[/latex]