Fractie reductibila ?
Moderators: Bogdan Posa, Laurian Filip
Fractie reductibila ?
Se da fractia \( \frac{7+5^{2n+1}\cdot2^{2n}}{4+5^{3n}\cdot2^{3n+1}} \), cu \( n \in N \). Stabiliti daca este reductiblia.
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Andi Brojbeanu
- Bernoulli
- Posts: 294
- Joined: Sun Mar 22, 2009 6:31 pm
- Location: Targoviste (Dambovita)
Pentru \( n=0: \frac{7+5^{2n+1}\cdot 2^{2n}}{4+5^{3n}\cdot 2^{3n+1}}=\frac{7+5}{4+2}=\frac{12}{6} \), fractie reductibila.
Pentru \( n\geq 1: \)\( \frac{7+5^{2n+1}\cdot 2^{2n}}{4+5^{3n}\cdot 2^{3n+1}} \)\( =\frac{7+5^{2n}\cdot 5\cdot 2^{2n}}{4+5^{3n}\cdot 2^{3n}\cdot 2} \)\( =\frac{7+5\cdot 10^{2n}}{4+2\cdot 10^{3n}} \)\( \frac{500.....007}{200......004} \).
Dar \( 7+0+0+...+0+0+5=12\Rightarrow \) numaratorul \( \vdots\ 3\ (1) \).
Dar \( 2+0+0+...+0+0+4=6\Rightarrow \) numitorul \( \vdots\ 3\ (2) \).
Din \( (1) \) si \( (2)\Rightarrow \) fractia este reductibila.
Pentru \( n\geq 1: \)\( \frac{7+5^{2n+1}\cdot 2^{2n}}{4+5^{3n}\cdot 2^{3n+1}} \)\( =\frac{7+5^{2n}\cdot 5\cdot 2^{2n}}{4+5^{3n}\cdot 2^{3n}\cdot 2} \)\( =\frac{7+5\cdot 10^{2n}}{4+2\cdot 10^{3n}} \)\( \frac{500.....007}{200......004} \).
Dar \( 7+0+0+...+0+0+5=12\Rightarrow \) numaratorul \( \vdots\ 3\ (1) \).
Dar \( 2+0+0+...+0+0+4=6\Rightarrow \) numitorul \( \vdots\ 3\ (2) \).
Din \( (1) \) si \( (2)\Rightarrow \) fractia este reductibila.