Question

\(A=\frac{5x-2}{x-2}\)

Tìm x thuộc tập hợp số nguyên để A có giá trị nguyên

Answer 1

\(A=\frac{5x-2}{x-2}=\frac{5\left(x-2\right)+10-2}{x-2}=\frac{5\left(x-2\right)}{x-2}+\frac{10-2}{x-2}=5+\frac{8}{x-2}\)

- Để A nguyên thì 8 phải chia hết cho x - 2

=>\(x-2\varepsilonƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

=> \(x\varepsilon\left\{-6;-2;0;1;3;4;6;10\right\}\)

Vậy:...

Answer 2

\(A=\frac{5x-2}{x-2}\)

\(=\frac{5x-10+8}{x-2}\)

\(=\frac{5\left(x-2\right)+8}{x-2}\)

\(=\frac{5\left(x-2\right)}{x-2}+\frac{8}{x-2}\)

\(=5+\frac{8}{x-2}\)

Để \(A\in Z\Rightarrow8⋮\left(x-2\right)\)

\(\Rightarrow x-2\in U\left(8\right)=\left\{1;2;4;8;-1;-2;-4;-8\right\}\)

\(\Rightarrow x=\left\{3;4;6;10;1;0;-2;-6\right\}\)