Question

Determinați elementele următoarelor mulțumi​

Answer 1

 

$\displaystyle\\ b)\\B=\left\{x\in N\Big|\frac{8}{x-2}\in N\right\}\\\\\frac{8}{x-2}\in N\implies8\vdots(x-2)\\\\D_8=\{1;~2;~4;~8\}\\\\x-2=1\implies x_1=1+2=3\\x-2=2\implies x_2=2+2=4\\x-2=4\implies x_3=4+2=6\\x-2=8\implies x_4=8+2=10\\\implies B=\{3;~4;~6;~10\}$

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$\displaystyle\\ c)\\C=\left\{x\in N\Big|\frac{3x-2}{x+4}\in N\right\}\\\\\frac{3x-2}{x+4}=\frac{3x+12-12-2}{x+4}=\frac{3x+12-14}{x+4}=\\\\=\frac{3x+12}{x+4}-\frac{14}{x+4}=\frac{3(x+4)}{x+4}-\frac{14}{x+4}=3-\frac{14}{x+4}\\\\\left(3-\frac{14}{x+4}\right)\in N\implies \frac{14}{x+4}\leq3\\\\D_{14}=\[1;~2;~7;~14\]~\text{din care excludem pe 1 si pe 2.}\\x+4=7\implies x_1=7-4=3\\x+4=14\implies x_2=14-4=10\\\implies C=\{3;~10\}$

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$\displaystyle\\d)\\D=\left\{x\in N\Big|\frac{4x+3}{5x-1}\in N \right\}\\\\\frac{4x+3}{5x-1}\in N~\text{ Rezulta ca si }~~\left(5\cdot \frac{4x+3}{5x-1}\right)\in N\\\\5\cdot \frac{4x+3}{5x-1}=\frac{5(4x+3)}{5x-1}=\frac{20x+15}{5x-1}\\\\\frac{20x-4+4+15}{5x-1}=\frac{20x-4+19}{5x-1}=\\\\=\frac{20x-4}{5x-1}+\frac{19}{5x-1}=\frac{4(5x-1)}{5x-1}+\frac{19}{5x-1}=4+\frac{19}{5x-1}\\\\D_{19}=\{1;~19\}\\5x-1=1\implies x=\frac{2}{5}\notin N\\5x-1=19\implies x=\frac{19+1}{5}=4\\\\\implies D=\{4\}$

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