prosze o rozwiazanie zad 2 c i zad 4 b i c. :) jak najszybciej :) dziekuje

zad 2 [latex][(3-sqrt{5})^{sqrt{2}}(2+sqrt{5})^{sqrt{2}}]^{sqrt{2}}=[((3-sqrt{5})(3+sqrt{5}))^{sqrt{2}}]^{sqrt{2}}=\ \ =(3^{2}-sqrt{2})^{sqrt{2}*sqrt{2}}=(9-5)^{2}=4^{2}=16\ -----\ Wzory:\ a^{n}*b^{n}=(a*b)^{n}\ (a-b)(a+b)=a^{2}+b^{2}\ (a^{n})^{m}=a^{n*m}\a^{n}*a^{m}=a^{n+m}\a^{n}:a^{m}=a^{n-m}[/latex] ================================================== zad 4 [latex]frac{3a^{5}s^{2}a+(a^{3})^{2}b^{2}}{0,5ba^{3}}=frac{3a^{6}s^{2}+a^{6}b^{2}}{frac{1}{2}ba^{3}}=frac{2a^{6}(3s^{2}+b^{2})}{ba^{3}}=frac{2a^{3}(3s^{2}+b^{2})}{b}\ ---------------\ frac{(a^{4})^{5}:(ab)^{8}}{(2a^{7}b^{3}-b^{3}a^{7}):b^{11}}=frac{a^{20}}{a^{8}b^{8}}:frac{2a^{7}b^{3}-a^{7}b^{3}}{b^{11}}=frac{a^{12}}{b^{8}}*frac{b^{11}}{a^{7}b^{3}}=frac{a^{5}}{1}*frac{b^{3}}{b^{3}}=a^{5}[/latex]