Question

9.Rezolvati in multimea umerelor intregi nenule ecuatia :3x+4y=24,|x|<12
10.Aratati ca fractia 7n+12 este ireductibila pentru orice n sa aprtina lui N
supra
3n+5

Answer 1

9. 3x+4y=24⇒2divide 24 si 2 divide 4y oricare ar fi y∈Z⇒2 divide 3x unde x-numar par.
|x|<12
x-numar par⇒x={10,8,6,4,3,2,-2,-4,-6,-8,-10}
10 si 8 nu convine deci x≠10;8.
Daca x=6⇒y=6/4∉Z⇒nu convine;
Daca x=4⇒y=3∈Z;
Daca x=2⇒y=18/4∉Z⇒nu convine;
Daca x=-2⇒y=30/4∉Z⇒nu convine;
Daca x=-4⇒y=9∈Z;
Daca x=-6⇒y=42/4∉Z⇒nu convine;
Daca x=-8⇒y=48/4=12∈Z;
Daca x=-10⇒y=54/4∈Z⇒nu convine;
Deci (x;y)={(4;3),(-4;9),(-8;12)};
10. Presupunem prin absurd faptul ca 7n+12/3n+5 este fractie reductibila
7n+12/3n+5=21n+36/21n+15⇒3n+5/36-15=21⇒3n+5∉D21={1;3;7;21}⇒contradictie fractia 7n+12/3n+5 este ireductibila;