Question

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

Answer 1

Đề là j z pn

Answer 2

Bổ sung đề:

Tìm \(x\in Z\) để \(A=\dfrac{2x+1}{x-3}\in Z\).

Lời giải:

Ta có: \(A=\dfrac{2x+1}{x-3}=\dfrac{\left(2x-6\right)+7}{x-3}=\dfrac{2x-6}{x-3}+\dfrac{7}{x-3}=2+\dfrac{7}{x-3}\)

Để \(A\in Z\) thì \(\dfrac{7}{x-3}\in Z\Rightarrow7⋮x-3\Leftrightarrow x-3\in U\left(7\right)=\left\{\pm1;\pm7\right\}\)

Ta có bảng sau:

\(x-3\)	\(1\)	\(-1\)	\(7\)	\(-7\)
\(x\)	\(4\)	\(2\)	\(10\)	\(-4\)

Vậy với \(x\in\left\{-4;2;4;10\right\}\) thì \(A=\dfrac{2x+1}{x-3}\in Z\)