Question

Sa se determine x natural astfel incat fractia (2x+13)/(x-4) să se simplifice cu 3 și 7

Answer 1

   
[tex]\displaystyle\\ \frac{2x+13}{x-4}~\vdots~3~~~\text{ si }~~~\frac{2x+13}{x-4}~\vdots~7\\\\ 3\times7=21\\\\ \Longrightarrow~~~\frac{2x+13}{x-4}~\vdots~21\\\\ \text{Rezolvare:}\\\\ 2(x-4)=2x-8~~~\text{unde }~(x-4)~\text{ este numitorul.}\\\\ \Longrightarrow~~\text{La numarator scadem 8 si adunam 8}~~(-8+8)\\\\ \Longrightarrow~~~\frac{2x+13}{x-4}=\frac{2x-8+8+13}{x-4}=\frac{2x-8+21}{x-4}=\\\\ =\frac{2x-8}{x-4}+\frac{21}{x-4}=\frac{2(x-4)}{x-4}+\frac{21}{x-4} [/tex]


[tex]\displaystyle\\ \left(\frac{2(x-4)}{x-4}+\frac{21}{x-4}\right) ~\vdots ~21 ~\texttt{ daca }~ x-4 = 21\cdot k\\\\\text{(unde 21k este un multiplu al lui 21)}\\\\ \text{Pentru }~k=1 ~\Longrightarrow~x-4=21 ~\Longrightarrow~x = \boxed{\bf25}\\\\ \text{Pentru }~k=n ~\Longrightarrow~x-4=21\cdot n ~\Longrightarrow~x = \boxed{\bf 21\cdot n +4};~n\in N \backslash \{0\}\\\\ \text{Verificare pentru: } ~x=25\\\\ \frac{2x+13}{x-4} = \frac{2\cdot 25+13}{25-4} =\frac{63}{21} ~\vdots ~21 [/tex]


[tex]\displaystyle\\ \text{Verificare pentru: } ~x=21\cdot n +4\\\\ \frac{2x+13}{x-4} = \frac{2\cdot (21\cdot n +4)+13}{(21\cdot n +4)-4} =\\\\ =\frac{42n+8+13}{21n+4-4} =\frac{42n+21}{21n} =\frac{21(2n+1)}{21n} ~\vdots ~21[/tex]

Answer 2

(3;7)=1⇒c.m.m.c(3;7)=3*7=21

2x+13=21k, k∈N
x-4=21p, p∈N

scadem a doua relatie din prima

2x+13-(x-4)=21k-21p
x+17=21(k-p) =21s
cum x∈N, 2x-13>x-4 deci k-p>0 , s∈N*

x=21s-17
cu conditia x∈N⇒s∈N*
x=4;25;46...21s-17 cu s∈N*
sau, aceeasi multime scrisa altfel,
 21m+4 cu m∈N

multimea solutiilor scrisa in 3 feluri echivalente este
x∈{21m+4|m∈N}={21s-17|s∈N*}={4;25;46...}