![\displaystyle{\frac{\sqrt[3]{7+5\sqrt{2}}\cdot (\sqrt{2}-1)}{\sqrt{4+2\sqrt{3}}-\sqrt{3}}}](/forum/ext/geomar/texintegr/latexrender/pictures/0078441948317f43b5fda475494cdbfd.png "\displaystyle{\frac{\sqrt[3]{7+5\sqrt{2}}\cdot (\sqrt{2}-1)}{\sqrt{4+2\sqrt{3}}-\sqrt{3}}}")
είναι ακέραιος.Είναι ακέραιος !
Συντονιστής: stranton
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parmenides51
- Δημοσιεύσεις: 6238
- Εγγραφή: Πέμ Απρ 23, 2009 9:13 pm
- Τοποθεσία: Πεύκη
- Επικοινωνία:
![\frac{{\sqrt[3]{{7 + 5\sqrt 2 }} \cdot \left( {\sqrt 2 - 1} \right)}}{{\sqrt {4 + 2\sqrt 3 } - \sqrt 3 }} =
\frac{{\sqrt[3]{{7 + 5\sqrt 2 }} \cdot \sqrt[3]{{{{\left( {\sqrt 2 - 1} \right)}^3}}}}}{{\sqrt {3 + 2\sqrt 3 + 1} - \sqrt 3 }} =](/forum/ext/geomar/texintegr/latexrender/pictures/85566579ee5f3d95fd2147c91388b2f8.png "\frac{{\sqrt[3]{{7 + 5\sqrt 2 }} \cdot \left( {\sqrt 2 - 1} \right)}}{{\sqrt {4 + 2\sqrt 3 } - \sqrt 3 }} =
\frac{{\sqrt[3]{{7 + 5\sqrt 2 }} \cdot \sqrt[3]{{{{\left( {\sqrt 2 - 1} \right)}^3}}}}}{{\sqrt {3 + 2\sqrt 3 + 1} - \sqrt 3 }} =")
![\frac{{\sqrt[3]{{7 + 5\sqrt 2 }} \cdot \sqrt[3]{{{{\left( {\sqrt 2 } \right)}^3} - 3 \cdot 2 \cdot 1 + 3\sqrt 2 - 1}}}}{{\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} - \sqrt 3 }} =](/forum/ext/geomar/texintegr/latexrender/pictures/56326c9cfca1082d1c91981c9576473f.png "\frac{{\sqrt[3]{{7 + 5\sqrt 2 }} \cdot \sqrt[3]{{{{\left( {\sqrt 2 } \right)}^3} - 3 \cdot 2 \cdot 1 + 3\sqrt 2 - 1}}}}{{\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} - \sqrt 3 }} =")
![\frac{{\sqrt[3]{{7 + 5\sqrt 2 }} \cdot \sqrt[3]{{2\sqrt 2 + 3\sqrt 2 - 7}}}}{{\sqrt 3 + 1 - \sqrt 3 }} =](/forum/ext/geomar/texintegr/latexrender/pictures/029c609a5ce718352e8a6720f55c4c20.png "\frac{{\sqrt[3]{{7 + 5\sqrt 2 }} \cdot \sqrt[3]{{2\sqrt 2 + 3\sqrt 2 - 7}}}}{{\sqrt 3 + 1 - \sqrt 3 }} =")
![\sqrt[3]{{\left( {7 + 5\sqrt 2 } \right)\left( {5\sqrt 2 - 7} \right)}} =](/forum/ext/geomar/texintegr/latexrender/pictures/acbe5d31853483c5d55bc34b8c501ba0.png "\sqrt[3]{{\left( {7 + 5\sqrt 2 } \right)\left( {5\sqrt 2 - 7} \right)}} =")
![\sqrt[3]{{50 - 49}} = 1](/forum/ext/geomar/texintegr/latexrender/pictures/1ae67ab7b37a5b7124b9343efb8b698a.png "\sqrt[3]{{50 - 49}} = 1")
![\displaystyle{\frac{\sqrt[3]{7+5\sqrt{2}}\cdot (\sqrt{2}-1)}{\sqrt{4+2\sqrt{3}}-\sqrt{3}}=\frac{\sqrt[3]{(1+\sqrt{2})^3}\cdot (\sqrt{2}-1)}{\sqrt{(1+\sqrt{3})^2}-\sqrt{3}}=(\sqrt{2}+1)(\sqrt{2}-1)=1}](/forum/ext/geomar/texintegr/latexrender/pictures/637fe764dd77dd0c75eb7bd415efb38f.png "\displaystyle{\frac{\sqrt[3]{7+5\sqrt{2}}\cdot (\sqrt{2}-1)}{\sqrt{4+2\sqrt{3}}-\sqrt{3}}=\frac{\sqrt[3]{(1+\sqrt{2})^3}\cdot (\sqrt{2}-1)}{\sqrt{(1+\sqrt{3})^2}-\sqrt{3}}=(\sqrt{2}+1)(\sqrt{2}-1)=1}")
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Γιώργος Ρίζος
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- Εγγραφή: Δευ Δεκ 29, 2008 1:18 pm
- Τοποθεσία: Κέρκυρα
Re: Είναι ακέραιος !
![\displaystyle
\begin{array}{l}
\sqrt[3]{{7 + 5\sqrt 2 }} \cdot \left( {\sqrt 2 - 1} \right) = \sqrt[3]{{2\sqrt 2 + 6 + 3\sqrt 2 + 1}} \cdot \left( {\sqrt 2 - 1} \right) = \\
\\
= \sqrt[3]{{\left( {\sqrt 2 } \right)^3 + 3 \cdot \left( {\sqrt 2 } \right)^2 \cdot 1 + 3\sqrt 2 \cdot 1 + 1^3 }} \cdot \left( {\sqrt 2 - 1} \right) = \\
\\
= \sqrt[3]{{\left( {\sqrt 2 + 1} \right)^3 }} \cdot \left( {\sqrt 2 - 1} \right) = \left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right) = 2 - 1 = 1 \\
\end{array}](/forum/ext/geomar/texintegr/latexrender/pictures/b0b1caf01a0e3387e490a2d50d295ff0.png "\displaystyle
\begin{array}{l}
\sqrt[3]{{7 + 5\sqrt 2 }} \cdot \left( {\sqrt 2 - 1} \right) = \sqrt[3]{{2\sqrt 2 + 6 + 3\sqrt 2 + 1}} \cdot \left( {\sqrt 2 - 1} \right) = \\
\\
= \sqrt[3]{{\left( {\sqrt 2 } \right)^3 + 3 \cdot \left( {\sqrt 2 } \right)^2 \cdot 1 + 3\sqrt 2 \cdot 1 + 1^3 }} \cdot \left( {\sqrt 2 - 1} \right) = \\
\\
= \sqrt[3]{{\left( {\sqrt 2 + 1} \right)^3 }} \cdot \left( {\sqrt 2 - 1} \right) = \left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right) = 2 - 1 = 1 \\
\end{array}")
οπότε
.edit: Ας πούμε ότι επεξήγησα την παραπάνω φωτεινή λύση!
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