Đáp án:
$a) x \in \{-2;0;-6;4\}\\ b)x \in \left\{0;2\right\}\\ c) x \in \left\{-5;-3;-1;1\right\}\\ d)x \in \left\{-4;-2;0;2\right\}$
Giải thích các bước giải:
$a) \dfrac{f(x)}{g(x)}\\ =\dfrac{2x^2-x+2}{x+1} \\ =\dfrac{2x^2+2x-3x-3+5}{x+1} \\ =\dfrac{2x(x+1)-3(x+1)+5}{x+1} \\ =2x-3+\dfrac{5}{x+1} $
Để $\dfrac{f(x)}{g(x)}$ nguyên với $x$ nguyên thì $\dfrac{5}{x+1}$ phải nguyên
$\Leftrightarrow \left\{\begin{array}{l} (x+1) \in Ư(5)\\ x \in \mathbb{Z} \end{array} \right.\\ \Rightarrow x \in \{-2;0;-6;4\}\\ b)\dfrac{f(x)}{g(x)}\\ =\dfrac{3x^2-4x+6}{ 3x-1} \\ =\dfrac{3x^2-x-3x+1+5}{ 3x-1} \\ =\dfrac{x(3x-1)-(3x-1)+5}{ 3x-1} \\ =x-1+\dfrac{5}{ 3x-1} $
Để $\dfrac{f(x)}{g(x)}$ nguyên với $x$ nguyên thì $\dfrac{5}{3x-1}$ phải nguyên
$\Leftrightarrow \left\{\begin{array}{l} (3x-1) \in Ư(5)\\ x \in \mathbb{Z} \end{array} \right.\\ \Rightarrow x \in \left\{0;2\right\}\\ c)\dfrac{f(x)}{g(x)}\\ =\dfrac{ -2x^3-7x^2-5x+5}{ x+2} \\ =\dfrac{ -2x^3-4x^2-3x^2-6x+x+2+3}{ x+2} \\ =\dfrac{ -2x^2(x+2)-3x(x+2)+x+2+3}{ x+2} \\ = -2x^2-3x+1+\dfrac{3}{ x+2} $
Để $\dfrac{f(x)}{g(x)}$ nguyên với $x$ nguyên thì $\dfrac{3}{ x+2}$ phải nguyên
$\Leftrightarrow \left\{\begin{array}{l} (x+2) \in Ư(3)\\ x \in \mathbb{Z} \end{array} \right.\\ \Rightarrow x \in \left\{-5;-3;-1;1\right\}\\ d)\dfrac{f(x)}{g(x)}\\ =\dfrac{ x^3-3x^2-4x+3 }{ x+1} \\ =\dfrac{ x^3+x^2-4x^2-4x+3 }{ x+1} \\ =\dfrac{ x^2(x+1)-4x(x+1)+3 }{ x+1} \\ = x^2-4x+\dfrac{3 }{ x+1} $
Để $\dfrac{f(x)}{g(x)}$ nguyên với $x$ nguyên thì $\dfrac{3}{ x+1}$ phải nguyên
$\Leftrightarrow \left\{\begin{array}{l} (x+1) \in Ư(3)\\ x \in \mathbb{Z} \end{array} \right.\\ \Rightarrow x \in \left\{-4;-2;0;2\right\}$
