Giải thích các bước giải:

$\begin{aligned}
& \text{a) } f(x) = 2x^2 - x + 2 \text{ ; } g(x) = x + 1 \\
& f(x) = 2x^2 + 2x - 3x - 3 + 5 \\
& f(x) = 2x(x + 1) - 3(x + 1) + 5 \\
& f(x) = (x + 1)(2x - 3) + 5 \\
& f(x) \vdots g(x) \\
& 5 \vdots (x + 1) \\
& x + 1 \in \{-5; -1; 1; 5\} \\
& x \in \{-6; -2; 0; 4\} \\
& \text{Kết quả: } x \in \{-6; -2; 0; 4\} \\
& \text{b) } f(x) = 3x^2 - 4x + 6 \text{ ; } g(x) = 3x - 1 \\
& f(x) = 3x^2 - x - 3x + 1 + 5 \\
& f(x) = x(3x - 1) - (3x - 1) + 5 \\
& f(x) = (3x - 1)(x - 1) + 5 \\
& f(x) \vdots g(x) \\
& 5 \vdots (3x - 1) \\
& 3x - 1 \in \{-5; -1; 1; 5\} \\
& 3x \in \{-4; 0; 2; 6\} \\
& x \in \mathbb{Z} \\
& x \in \{0; 2\} \\
& \text{Kết quả: } x \in \{0; 2\} \\
& \text{c) } f(x) = -2x^3 - 7x^2 - 5x + 5 \text{ ; } g(x) = x + 2 \\
& f(x) = -2x^3 - 4x^2 - 3x^2 - 6x + x + 2 + 3 \\
& f(x) = -2x^2(x + 2) - 3x(x + 2) + (x + 2) + 3 \\
& f(x) = (x + 2)(-2x^2 - 3x + 1) + 3 \\
& f(x) \vdots g(x) \\
& 3 \vdots (x + 2) \\
& x + 2 \in \{-3; -1; 1; 3\} \\
& x \in \{-5; -3; -1; 1\} \\
& \text{Kết quả: } x \in \{-5; -3; -1; 1\}
\end{aligned}$