Question

[TranNgocThienThu]

Tìm tất cả các số nguyên a biết : \(\left(6a+1\right)⋮\left(3a-1\right)\)

Answer 1

[Dương Lam Hàng]

Ta có: \(6a+1=2\left(3a-1\right)+3\)

Vì \(2\left(3a-1\right)⋮\left(3a-1\right)\Rightarrow3⋮\left(3a-1\right)\)

\(\Rightarrow3a-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

Nếu 3a - 1 = 1 thì 3a = 2 => a = 2/3

Nếu 3a - 1 = -1 thì 3a = 0 => a = 0

Nếu 3a - 1 = 3 thì 3a = 4 => a = 4/3

Nếu 3a - 1 = -3 thì 3a = -2 => a = -2/3

Mà \(a\in Z\Rightarrow a=0\)

Vậy a = 0

Answer 2

[KAl(SO4)2·12H2O]

\(\left(6a+1\right)⋮\left(3a-1\right)\)

\(\Rightarrow\left[\left(6a+1\right)-2\left(3a-1\right)\right]⋮\left(3a-1\right)\)

\(\Rightarrow\left(6a+1-6a+2\right)⋮\left(3a-1\right)\)

\(\Rightarrow3⋮\left(3a-1\right)\)

\(\Rightarrow3a-1\inƯ\left(3\right)\)

\(\Rightarrow3a-1\in\left\{\pm1;\pm3\right\}\)

\(\Rightarrow3a\in\left\{\pm2;0;4\right\}\)

\(\Rightarrow a\in\left\{\frac{2}{3};0;\frac{4}{3};-\frac{2}{3}\right\}\)

Mà \(a\in Z\)

\(\Rightarrow a=0\)