Ta có : 2x - 1 = 2x - 6 + 5 = (2x - 6) + 5 = 2 . (x - 3) + 5

Vì x - 3 chia hết cho x - 3 nên 2 . (x - 3) chia hết cho x - 3

Suy ra , 5 phải chia hết cho x - 3

Hay \(x-3\in\left\{1;-1;5;-5\right\}\)

\(\Rightarrow x\in\left\{4;2;8;-2\right\}\)

Mà x là số nguyên dương nên \(x\in\left\{2;4;8\right\}\)

Vậy \(x\in\left\{2;4;8\right\}\)

_HT_

\(\dfrac{2\left(x-3\right)+5}{x-3}=2+\dfrac{5}{x-3}\Rightarrow x-3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(2\left(x-3\right)+5⋮x-3\Rightarrow x-3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(=x\in\left\{2;4;8\right\}\)

ta có: \(2x-1=2\left(x-3\right)+5\)

để \(2x-1⋮x-3\Rightarrow2\left(x-3\right)+5⋮x-3\\ m\text{à }x.nguy\text{ê}n\Rightarrow x-3nguy\text{ê}n\\ \Rightarrow x-3\in\text{Ư}\left(5\right)=\left\{-5;5;1;-1\right\}\)

ta có bảng sau :

\(\Leftrightarrow2.\left(x-3\right)+5⋮x-3\)

\(do2.\left(x-3\right)⋮x-3\)

\(\Leftrightarrow5⋮x-3\)

\(\Leftrightarrow x-3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)

\(\Leftrightarrow x\in\left\{-2;2;4;8\right\}\)

(a) \(f\left(x\right)⋮g\left(x\right)\Rightarrow\dfrac{x^2-5x+9}{x-3}\in Z\)

Ta có: \(\dfrac{x^2-5x+9}{x-3}\left(x\ne3\right)=\dfrac{x\left(x-3\right)-2\left(x-3\right)+3}{x-3}=x-2+\dfrac{3}{x-3}\)nguyên khi và chỉ khi: \(\left(x-3\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\\x-3=3\\x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\\x=6\\x=0\end{matrix}\right.\) (thỏa mãn).

Vậy: \(x\in\left\{0;2;4;6\right\}\).

(b) \(f\left(x\right)⋮g\left(x\right)\Rightarrow\dfrac{2x^3-x^2+6x+2}{2x-1}\in Z\left(x\ne\dfrac{1}{2}\right)\)

Ta có: \(\dfrac{2x^3-x^2+6x+2}{2x-1}=\dfrac{x^2\left(2x-1\right)+3\left(2x-1\right)+5}{2x-1}=x^2+3+\dfrac{5}{2x-1}\)

nguyên khi và chỉ khi: \(\left(2x-1\right)\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=1\\2x-1=-1\\2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\\x=3\\x=-2\end{matrix}\right.\) (thỏa mãn).

Vậy: \(x\in\left\{-2;0;1;3\right\}\).

a: f(x) chia hết cho g(x)

=>x^2-3x-2x+6+3 chia hết cho x-3

=>3 chia hết cho x-3

=>x-3 thuộc {1;-1;3;-3}

=>x thuộc {4;2;6;0}

b: f(x) chia hết cho g(x)

=>2x^3-x^2+6x-3+5 chia hết cho 2x-1

=>5 chia hết cho 2x-1

=>2x-1 thuộc {1;-1;5;-5}

=>x thuộc {2;0;3;-2}