Question

Sa se determine multimea A={x apartine Z cu proprietatea ca (2x³-3x²+4x+9)/2x+1 apartine Z}

Answer 1

Am atasat rezolvarea:

Answer 2

$\displaystyle 2x^3-3x^2+4x+9= \\ \\ =2x^3+x^2-4x^2-2x+6x+9= \\ \\ =x^2(2x+1)-2x(2x+1)+6x+9= \\ \\ =(2x+1)(x^2-2x)+6x+9. \\ \\ Deci~ \frac{2x^3-3x^2+4x+9}{2x+1}= \frac{(2x+1)(x^2-2x)+6x+9}{2x+1}= \\ \\ =x^2-2x+ \frac{6x+9}{2x+1} \in \mathbb{Z} \Leftrightarrow 2x+1~|~6x+9.$

$\displaystyle 2x+1~|~2x+1 \Rightarrow 2x+1~|~6x+3. \\ \\ Deci~2x+1~|~(6x+9)-(6x+3) \Leftrightarrow 2x+1~|~6. \\ \\ Deci~2x+1 \in D_6,~si~cum~2x+1~este~impar,~rezulta \\ \\ 2x+1 \in \{-3,-1,1,3 \}. \\ \\ Prin~urmare~x \in \{-2,-1,0,1 \}. \\ \\ A=\{-2,-1,0,1 \}.$