Question

Cho A = \(\dfrac{3x^2-x+1}{3x+2}\) ( x khác \(\dfrac{-2}{3}\))

Tìm x thuộc Z để A thuộc Z

Answer 1

Để A thuộc Z thì \(3x^2-x+1⋮3x+2\)

\(3x^2+2x-3x-2+3⋮3x+2\)

\(x\left(3x+2\right)-\left(3x+2\right)+3⋮3x+2\)

\(\left(3x+2\right)\left(x-1\right)+3⋮3x+2\)

Mà \(\left(3x+2\right)\left(x-1\right)⋮3x+2\)

\(\Rightarrow3⋮3x+2\)

\(\Rightarrow3x+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Ta có bảng :

3x+2	1	-1	3	-3
x	-1/3	-1	1/3	-5/3

Mà x thuộc Z => x = -1

Vậy x = -1

Answer 2

\(A=\dfrac{3x^2+2x-3x+1}{3x+2}=\dfrac{3x^2+2x-3x-2+3}{3x+2}\)

\(A=\dfrac{x\left(3x+2\right)-\left(3x+2\right)+3}{3x+2}=x-1+\dfrac{3}{3x+2}\in Z\)

\(\Rightarrow3x+2\inƯ\left(3\right)\)

Xét ước thôi