Luciaalinaask
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Se considera multimea A={-2,5; [tex] \sqrt\frac{4}{9}[/tex];[tex] \sqrt{2} [/tex];[tex] \sqrt{6,25} [/tex];[tex] \sqrt{ \frac{25}{3} } [/tex]} . Calculati A ∩ Q (multimea numerelor rationale)

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Matepentrutotiask
Stabilim intai numerele rationale din multimea A.
[tex]-2,5= \frac{-25}{10} \in Q\\ \sqrt{\frac{4}{9}}= \frac{ \sqrt{4} }{ \sqrt{9} } = \frac{2}{3} \in Q\\ \sqrt{2}\notin Q\\ \sqrt{6,25} = \sqrt{\frac{625}{100}}= \frac{25}{10}\in Q\\ \sqrt{ \frac{25}{3} } = \frac{5}{ \sqrt{3} } = \frac{5}{3} \sqrt{3} \notin Q\\ A\cap Q=\{-2,5;\sqrt{\frac{4}{9}}; \sqrt{6,25}\}[/tex]