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√3-√5 dar cu lamurire

Răspuns :

[tex] \sqrt{a- \sqrt{b} } = \sqrt{ \frac{a+c}{2} } -\sqrt{ \frac{a-c}{2} },\ unde\ c^2=b^2-a\\ c^2=3^2-5=>c=2\\ \sqrt{3- \sqrt{5} } = \sqrt{ \frac{3+2}{2} } -\sqrt{ \frac{3-2}{2} }=\sqrt{ \frac{5}{2} } -\sqrt{ \frac{1}{2} }= \frac{ \sqrt{5}-1 }{ \sqrt{2} } [/tex]