zad Oblicz: c) [latex] sqrt{ sqrt{7 } - sqrt{3} } [/latex] x [latex] sqrt{ sqrt{7}+ sqrt{3} } [/latex] d) [latex] sqrt{4- 2sqrt{3} } [/latex] x [latex] sqrt{4+2sqrt{3} } [/latex] x- znaczy razy

[latex] sqrt{ sqrt{7 } - sqrt{3} } cdot sqrt{ sqrt{7}+ sqrt{3} } = \ \ sqrt{( sqrt{7 } - sqrt{3}) cdot ( sqrt{7}+ sqrt{3}) } = \ \ sqrt{( sqrt{7 })^2 -( sqrt{3}) ^2 } = \ \ sqrt{7-3 } = sqrt4=2[/latex] ====================== [latex] sqrt{4- 2sqrt{3} }cdot sqrt{4+2sqrt{3} } = \ \ sqrt{(4- 2sqrt{3})cdot (4+2sqrt{3}) } = \ \ sqrt{4^2- (2sqrt{3})^2} = \ \ sqrt{16-12} =sqrt4=2[/latex]