Question

Tìm \(x\in Z\)để \(\frac{5-3x}{2x-1}\in Z\)

Answer 1

\(\frac{5-3x}{2x-1}\in Z\Rightarrow\frac{10-6x}{2x-1}\in Z\)

\(\frac{10-6x}{2x-1}=\frac{10-6x+3-3}{2x-1}=\frac{7-3\left(2x-1\right)}{2x-1}=\frac{7}{2x+1}-3\)

Để \(\frac{7}{2x-1}-3\in Z\Leftrightarrow\frac{7}{2x-1}\in Z\)

=> 2x - 1 ∈ Ư(7) = { ± 1; ± 7 }

Ta có : 2x - 1 = 7 => 2x = 8 => x = 4

           2x - 1 = 1 => 2x = 2 => x = 1

           2x - 1 = - 1 => 2x = 0 => x = 0

           2x - 1 = - 7 => 2x = - 6 => x = - 3

Vậy x = { - 3; 0; 1; 4 }