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 :

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

\(do2.\left(x-3\right)⋮x-3\)

\(\Leftrightarrow5⋮x-3\)

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

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

\(\Leftrightarrow2x^3-3x^2+x+a=\left(x+3\right)\cdot a\left(x\right)\)

Thay \(x=-3\)

\(\Leftrightarrow2\left(-27\right)-3\cdot9-3+a=0\\ \Leftrightarrow-54-27-3+a=0\\ \Leftrightarrow-84+a=0\\ \Leftrightarrow a=84\)

\(2x-1⋮x-3\)

\(=>2.\left(x-3\right)+5⋮x-3\)

Do \(2.\left(x-3\right)⋮x-3\)

\(=>5⋮x-3\)

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

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

TL :

Ô bỏ các số âm đi nhé

Vì đây là nguyên dương

HT

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_

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