Fractie ireductibila

alex2008 · Post by **alex2008** » Fri Jan 16, 2009 2:41 pm

Arãtaţi cã fracţia : \( \frac{5n+7}{6n^2+17n+12} \) este ireductibila \( \forall n \in \mathb{N} \) .

mihai miculita · Post by **mihai miculita** » Thu Mar 26, 2009 10:05 pm

\( \left \array (5n+7)\vdots d \\
(6n^2+17n+12)\vdots d \right}\Rightarrow 5.(6n^2+17n+12)-6n.(5n+7)=(43n+60)\vdots d \)
\( \left \array (5n+7)\vdots d \\
(43n+60)\vdots d \right}\Rightarrow 43.(5n+7)-5.(43n+60)=1\vdots d \Rightarrow (5n+7;6n^2+17n+12)=1; \ (\forall)n\in\mathbb{N}. \)