Strona 1 z 1
a la Viete
: 25 maja 2025, o 04:15
autor: Eariu52
\(\displaystyle{ \frac{2}{ \sqrt{3} }= \sqrt{2} \cdot \sqrt{2- \sqrt{2} } \cdot \sqrt{2- \sqrt{2- \sqrt{2} } } \cdot \ldots }\)
Re: a la Viete
: 25 wrz 2025, o 05:20
autor: Eariu52
\(\displaystyle{ \frac{2 \sqrt{2} }{3}=\left( \sqrt{2+ \sqrt{2} } - \sqrt{2- \sqrt{2} }\right) \cdot \left( \sqrt{2+ \sqrt{2+ \sqrt{2} } } - \sqrt{2- \sqrt{2- \sqrt{2} } }\right)\cdot \ldots }\)
Re: a la Viete
: 8 gru 2025, o 04:00
autor: Eariu52
\(\displaystyle{ \sqrt{3} = \frac{ \sqrt{2}+1 }{ \sqrt{2} } \cdot \frac{ \sqrt{2- \sqrt{2} }+1 }{ \sqrt{2+ \sqrt{2} } } \cdot \frac{ \sqrt{2- \sqrt{2- \sqrt{2} } }+1 }{ \sqrt{2+ \sqrt{2+ \sqrt{2} } } } \cdot \ldots}\)