theo mình nghĩ đây là tìm GTNN 

19 chia hết cho x

x \(\in\) Ư_{( 19 )}

x \(\in\) { 1 ; 19 }

a) Ta có: \(19⋮x\)

\(\Leftrightarrow x\inƯ\left(19\right)\)

hay \(x\in\left\{1;-1;19;-19\right\}\)

Vậy: \(x\in\left\{1;-1;19;-19\right\}\)

b) Ta có: \(23⋮x+1\)

\(\Leftrightarrow x+1\inƯ\left(23\right)\)

\(\Leftrightarrow x+1\in\left\{1;-1;23;-23\right\}\)

hay \(x\in\left\{0;-2;22;-24\right\}\)

Vậy: \(x\in\left\{0;-2;22;-24\right\}\)

c) Ta có: \(12⋮x-1\)

\(\Leftrightarrow x-1\inƯ\left(12\right)\)

\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

hay \(x\in\left\{2;0;3;-1;4;-2;5;-3;7;-5;13;-11\right\}\)

Vậy: \(x\in\left\{2;0;3;-1;4;-2;5;-3;7;-5;13;-11\right\}\)

x ∈ Z

a. 19 chia hết cho x

=> x ∈ \(\left\{-19;-1;1;19\right\}\)

b. 23 chia hết cho x+1

=> x+ 1 ∈ \(\left\{-23;-1;1;23\right\}\)

=> x ∈ \(\left\{-24;-2;0;22\right\}\)

c. 12 chia hết cho x-1

=> x-1 ∈ \(\left\{-12;-4;-3;-1;1;3;4;12\right\}\)

=> x ∈ \(\left\{-11;-3;-2;0;2;4;5;13\right\}\)

Dạng này của easy thôi!

\(A=\frac{x^2+4x-19}{x-3}=\frac{\left(x-3\right)^2}{x-3}+\frac{10x-28}{x-3}\)

\(=x-3+\frac{10\left(x-3\right)}{x-3}+\frac{2}{x-3}=x+7+\frac{2}{x-3}\)

Để A nguyên thì \(\frac{2}{x-3}\) nguyên tức là \(x-3\inƯ\left(2\right)\)

Giải ra tiếp!

Ta có: \(\frac{x^2+4x-19}{x-3}=\frac{\left(x^2-6x+9\right)+10x-28}{x-3}\)

\(=\frac{\left(x-3\right)^2}{x-3}+\frac{10x-28}{x-3}=x-3+\frac{10x-30+2}{x-3}\)

\(=x-3+\frac{10\left(x-3\right)+2}{x-3}=x-3+\frac{10\left(x-3\right)}{x-3}+\frac{2}{x-3}\)

\(=x-3+10+\frac{2}{x-3}=x+7+\frac{2}{x-3}\)

Để A là 1 số nguyên thì 2 \(⋮\) x - 3 => x - 3 \(\in\) Ư(2) = {-2; -1; 1; 2}
Nếu x - 3 = -2 => x = -2 + 3 => x = 1

Nếu x - 3 = -1 => x = -1 + 3 => x = 2

Nếu x - 3 = 1 => x = 1 + 3 => x = 4

Nếu x - 3 = 2 => x  = 2 + 3 => x = 5

Vậy x \(\in\) {1; 2; 4; 5}