Proszę o rozwiązanie tych dwóch zadań z obliczeniami . Daje Naj . z Góry Dziękuję

[latex]\9. \ \ dfrac{1(1+ sqrt{2}) }{(1- sqrt{2})(1+ sqrt{2} ) } = dfrac{1+ sqrt{2} }{1-2} =-1- sqrt{2} \ \Odp. B \ \10. \ \ sqrt{4+2 sqrt{3} } cdot sqrt{4-2 sqrt{3} } = sqrt{4^2-(2 sqrt{3})^2}= sqrt{16-12} =2 \ \Odp. A.[/latex]

[latex]9.\\frac{1}{1-sqrt{2}}cdotfrac{1+sqrt{2}}{1+sqrt{2}} = frac{1+sqrt{2}}{1-2} = frac{1+sqrt{2}}{-1} = -1-sqrt{2}\\Odp. B.[/latex] [latex]10.\\a = sqrt{4+2sqrt{3}}cdotsqrt{4-2sqrt{3}}=sqrt{(4+2sqrt{3})(4-2sqrt{3})}=\\=sqrt{4^{2}-(2sqrt{3})^{2}} = sqrt{16-12} = sqrt{4} = 2\\Odp. A.[/latex]