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usuwanie niewymiernosci

: 21 wrz 2012, 20:20
autor: moniaw094
\(\frac{1}{2 + \sqrt{5} + 2 \sqrt{2} + \sqrt{10} }\)

: 21 wrz 2012, 20:44
autor: irena
\(2+2\sqrt{2}+\sqrt{5}+\sqrt{10}=2(1+\sqrt{2})+\sqrt{5}(1+\sqrt{2})=(1+\sqrt{2})(2+\sqrt{5})\)

\(\frac{1}{2+\sqrt{5}+2\sqrt{2}+\sqrt{10}}=\frac{1}{(\sqrt{12}+1)(\sqrt{5}+2)}\cdot\frac{(\sqrt{2}-1)(\sqrt{5}-2)}{(\sqrt{2}-1)(\sqrt{5}-2)}=\frac{(\sqrt{2}-1)(\sqrt{5}-2)}{(2-1)(5-4)}=(\sqrt{2}-1)(\sqrt{5}-2)=\\=\sqrt{10}-2\sqrt{2}-\sqrt{5}+2\)

Re: usuwanie niewymiernosci

: 21 wrz 2012, 20:48
autor: wsl1993_
\(\frac{1}{2 + \sqrt{5} + 2 \sqrt{2} + \sqrt{10} }=\frac{1}{2 + \sqrt{5} + 2 \sqrt{2} + (\sqrt{2} \cdot\sqrt{5}) }
=\frac{1}{2 + 2 \sqrt{2} + \sqrt{5}+ (\sqrt{2} \cdot\sqrt{5}) }=\frac{1}{2(1+ \sqrt{2}) + \sqrt{5}(1+ \sqrt{2} ) }
=\frac{1}{(1+ \sqrt{2}) (sqrt5+2)} \cdot \frac{{(1- \sqrt{2}) (sqrt5-2)}}{(1- \sqrt{2}) (sqrt5-2)}
=\frac{(1- \sqrt{2}) (sqrt5-2)}{(1-2) (5-4)}=\frac{(\sqrt{5}-2- sqrt10+2\sqrt2)}{-1}=2-2\sqrt2-\sqrt5+\sqrt10
=2-2\sqrt2-\sqrt5+(sqrt5 \cdot sqrt2)=2(1-\sqrt2)-\sqrt5(1-\sqrt2)=(1-\sqrt2)(2-\sqrt5)\)