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

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

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

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_

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

\(x^3-3x^2-3x-1=\left(x-4\right)\left(x^2+x+1\right)+3\)

\(\Rightarrow x^3-3x^2-3x-1\) chia hết \(x^2+x+1\) khi \(3⋮x^2+x+1\)

\(\Rightarrow x^2+x+1=Ư\left(3\right)\) (1)

Mà x nguyên dương \(\Rightarrow x^2+x+1\ge1^2+1+1=3\) (2)

(1);(2) \(\Rightarrow x^2+x+1=3\)

\(\Rightarrow x=1\)